Ncert Solution for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

Ray Optics and Optical Instruments ncert solutions: Ncert Class 12 Physics Chapter 9 Exercise Solutions

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Question 9.1: A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

Solution 9.1: To solve this problem, we can use the mirror formula:

\(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)

where:

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• (f) is the focal length of the mirror,
• (v) is the image distance (where the screen needs to be placed),
• (u) is the object distance (the position of the candle).

Step 1: Determine the focal length

The focal length (f) of a concave mirror is related to the radius of curvature (R) by the formula:

\(f = \frac{R}{2}\)

Given R = 36 cm, we find:

\(f = \frac{36}{2} = 18 \, \text{cm}\)

Step 2: Use the mirror formula

The object distance (u) is given as -27 cm (negative because the object is in front of the mirror). Now, we can substitute the values into the mirror formula to find (v):

\(\frac{1}{18} = \frac{1}{v} + \frac{1}{-27}\)

Rearranging to solve for \(\frac{1}{v}\):

\(\frac{1}{v} = \frac{1}{18} + \frac{1}{27}\)

To add the fractions, we get a common denominator:

\(\frac{1}{v} = \frac{3}{54} + \frac{2}{54} = \frac{5}{54}\)

So,

\(v = \frac{54}{5} = -54 \, \text{cm}\)

Step 3: Nature and size of the image

• The negative sign for (v) indicates that the image is real (since it’s on the same side as the object for a concave mirror) and inverted.
• To find the magnification (m), we use the formula:

\(m = \frac{v}{u} = \frac{-54}{-27} = 2\)

The magnification is 2, meaning the image is twice the size of the object.

• The size of the object is 2.5 cm, so the size of the image is:

Image size = m × object size = 2 × 2.5 = 5.0cm

Thus, the image is real, inverted, and magnified, with a size of 5.0 cm.

If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.

Question 9.2: A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Solution 9.2: Step 1: Use the mirror formula

The mirror formula is:

\(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)

where:

• f is the focal length of the mirror,
• v is the image distance,
• uu is the object distance (the distance of the needle from the mirror).

For a convex mirror:

• The focal length (f) is positive, so f = +15 cm,
• The object distance u = -12 cm (negative because the object is in front of the mirror).

A convex mirror always form virtual image, which is erect and small in size of an object kept in front of it. Focal length of convex mirror ƒ = + 15 cm Object distance u = – 12 cm Using mirror formula

Substitute the values into the mirror formula:

\(\frac{1}{15} = \frac{1}{v} + \frac{1}{-12}\)

Rearranging to solve for \(\frac{1}{v}\):

\(\frac{1}{v} = \frac{1}{15} + \frac{1}{12}\)

To add the fractions, find a common denominator:

\(\frac{1}{v} = \frac{4}{60} + \frac{5}{60} = \frac{9}{60}\)

So,

\(v = \frac{60}{9} \approx 6.67 \, \text{cm}\)

Step 2: Determine the magnification

The magnification (m) is given by:

\(m = \frac{v}{u}\)

Substitute the values of v = 6.67 cm and u = -12 cm:

\(m = \frac{6.67}{-12} \approx -0.56\)

The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.

If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.

Question 9.3: A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

Solution 9.3: Step 1: Refractive Index of Water

The refractive index (n) of water can be calculated using the formula relating real depth, apparent depth, and refractive index:

\(n = \frac{\text{real depth}}{\text{apparent depth}}\)

Given:

• Real depth = 12.5 cm,
• Apparent depth = 9.4 cm.

Substitute the values into the formula:

\(n = \frac{12.5}{9.4} \approx 1.33\)

Thus, the refractive index of water is approximately 1.33.

Step 2: Change in Focus with the Liquid of Refractive Index 1.63

If the water is replaced by a liquid of refractive index 1.63, we need to calculate the new apparent depth and how much the microscope needs to be moved.

The formula for the refractive index is the same:

\(n_{\text{liquid}} = \frac{\text{real depth}}{\text{new apparent depth}}\)

Given:

• Refractive index of the new liquid, \(n_{\text{liquid}} = 1.63\),
• Real depth = 12.5 cm.

Rearranging the formula to find the new apparent depth:

new apparent depth = \(\frac{\text{real depth}}{n_{\text{liquid}}} = \frac{12.5}{1.63}\)

Calculating:

\(\text{new apparent depth} \approx 7.67 \, \text{cm}\)

Step 3: Movement of the Microscope

The microscope was initially focused at the apparent depth of 9.4 cm (for water). Now, the new apparent depth is 7.67 cm (for the new liquid). The distance the microscope needs to be moved is:

distance to move = 9.4 cm – 7.67 cm = 1.73 cm

Conclusion

• The refractive index of water is approximately 1.33.
• If the water is replaced with a liquid of refractive index 1.63, the microscope needs to be moved 1.73 cm upwards to focus on the needle again.

Question 9.4: Figures 9.27(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Fig. 9.27(c)].

Solution 9.4: For the glass-air interface:

The angle of incidence, i = 60°

The angle of refraction, r = 35°

From Snell’s law, we know that the refractive index of the glass with respect to air is given as

\(\mu_{g}^{a}=\frac{sin\;i}{sin\;r}\ \)

\(=\frac{sin\;60°}{sin\;35°}\)

\(=\frac{0.8660}{0.5736}=1.51\) …..(i)

For the air-water interface:

The angle of incidence, i = 60°

The angle of refraction, r = 47°

From Snell’s law, we know that the refractive index of water with respect to air is given as

\(\mu_{w}^{a}=\frac{sin\;i}{sin\;r}\ \)

\(=\frac{sin\;60°}{sin\;47°}\)

\(=\frac{0.8660}{0.7314}\)

\(=1.184\) …..(ii)

With the help of equations (i) and (ii), the relative refractive index of glass with respect to water can be found.

\(\mu_{g}^{w}=\frac{\mu_{g}^{a}}{\mu_{w}^{a}}\ \)

\(=\frac{1.51}{1.184}=1.275\)

The angle of incidence, i = 47°

The angle of refraction= r

r can be calculated from Snells’s law as

\(\frac{sin\;i}{sin\;r}=\mu_{g}^{w}\ \)

\(\frac{sin45°}{sin\;r}=1.275\ \)

\(sin\;r=\frac{\frac{1}{\sqrt{2}}}{1.275}=0.5546\ \)

Therefore r = \(sin^{-1}(0.5546)=38.68°\)

Hence, the angle of refraction at the water − glass interface is 38.68°.

Question 9.5: A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

Solution 9.5: This problem deals with the concept of total internal reflection and the critical angle. Light from the bulb at the bottom of the tank can only emerge from the water surface if its angle of incidence (inside the water) is less than the critical angle.

Step 1: Critical Angle Calculation

The critical angle \((\theta_c)\) is the angle of incidence beyond which light undergoes total internal reflection and does not escape the water. The critical angle can be calculated using the formula:

\(\sin \theta_c = \frac{1}{n}\)

where:

• (n) is the refractive index of water, (n = 1.33),
• The refractive index of air is approximately 1.

Substituting the value of (n):

\(\sin \theta_c = \frac{1}{1.33} \approx 0.7519\)

Now, calculate the critical angle:

\(\theta_c = \sin^{-1}(0.7519) \approx 48.75^\circ\)

Step 2: Geometry of the Situation

Light emerging from the bulb at the bottom of the tank forms a cone with the critical angle \((\theta_c)\). Only light rays within this cone can escape through the water’s surface.

• The bulb is at the bottom of the tank, so the distance from the bulb to the water surface is 80 cm (or 0.8 m).
• The radius of the circular area on the surface, through which light can emerge, can be found using trigonometry:

\(\text{tan} \, \theta_c = \frac{r}{h}\)

where:

• (r) is the radius of the circular area,
• (h = 80 cm = 0.8 m) is the depth of the water.

Rearranging to solve for (r):

\(r = h \times \tan \theta_c\)

Substitute the values:

\(r = 0.8 \times \tan(48.75^\circ)\)

Calculate:

\(r \approx 0.8 \times 1.139 = 0.911 \, \text{m}\)

Step 3: Area of the Surface Through Which Light Emerges

The area of the surface through which light can emerge is the area of the circle with radius (r). The area (A) of a circle is given by:

\(A = \pi r^2\)

Substitute (r = 0.911 m):

\(A \approx \pi \times (0.911)^2 \)

\(\approx 3.1416 \times 0.830 \)

\(\approx 2.61 \, \text{m}^2\)

The area of the surface of the water through which light from the bulb can emerge is approximately 2.61 m².

Question 9.6: A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Solution 9.6: To solve this problem, we need to use the prism formula and some principles of optics.

Step 1: Find the Refractive Index of the Prism Material

The formula for the angle of minimum deviation D_m for a prism is given by:

\(n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\)

where:

• (n) is the refractive index of the prism material,
• (A) is the angle of the prism,
• (D_m) is the angle of minimum deviation.

Given:

• Angle of minimum deviation, (D_m = 40°),
• Angle of the prism, (A = 60°).

Substitute these values into the formula:

\(n = \frac{\sin\left(\frac{60^\circ + 40^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)}\)

Calculate:

\(\frac{60^\circ + 40^\circ}{2} = 50^\circ\)

\(\frac{60^\circ}{2} = 30^\circ\)

So,

\(n = \frac{\sin 50^\circ}{\sin 30^\circ}\)

The sine values are:

\(\sin 50^\circ \approx 0.766\)

\(\sin 30^\circ = 0.5\)

Thus:

\(n = \frac{0.766}{0.5} \approx 1.532\)

Step 2: Calculate the New Angle of Minimum Deviation in Water

When the prism is placed in water, the effective refractive index of the prism material relative to water must be used. The formula to find the new angle of minimum deviation D’_m in water is:

\(n_{\text{eff}} = \frac{n_{\text{prism}}}{n_{\text{water}}}\)

where:

• (n_prism = 1.532),
• (n_water = 1.33).

Calculate the effective refractive index:

\(n_{\text{eff}} = \frac{1.532}{1.33} \approx 1.15\)

Now, use this effective refractive index in the prism formula to find the new angle of minimum deviation. Rearranging the prism formula to solve for (D_‘m).

Substitute:

\(\sin\left(\frac{60^\circ + D’_m}{2}\right) = 1.15 \cdot \sin 30^\circ\)

\(\sin\left(\frac{60^\circ + D’_m}{2}\right) = 1.15 \cdot 0.5 = 0.575\)

\(\frac{60^\circ + D’_m}{2} = \sin^{-1}(0.575)\)

Calculate:

\(\sin^{-1}(0.575) \approx 35.1^\circ\)

\(\frac{60^\circ + D’_m}{2} = 35.1^\circ\)

\(60^\circ + D’_m = 70.2^\circ\)

\(D’_m = 70.2^\circ – 60^\circ = 10.2^\circ\)

• The refractive index of the prism material is approximately 1.532.
• When the prism is placed in water, the new angle of minimum deviation of a parallel beam of light is approximately 10.2°.

Question 9.7: Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm?

Solution 9.7: To find the radius of curvature required for a double-convex lens with both faces having the same radius of curvature, and a given focal length, we can use the lens maker’s formula.

Lens Maker’s Formula

For a double-convex lens, the lens maker’s formula is:

\(\frac{1}{f} = (n – 1) \left( \frac{1}{R_1} – \frac{1}{R_2} \right)\)

where:

• (f) is the focal length of the lens,
• (n) is the refractive index of the lens material,
• (R₁) and (R₂) are the radii of curvature of the two lens surfaces.

For a double-convex lens with both faces having the same radius of curvature (R), and noting that ((R₁ = R) and (R₂ = -R) (the second surface is convex, so it is negative in sign convention), the formula simplifies to:

\(\frac{1}{f} = (n – 1) \left( \frac{1}{R} – \frac{(-1)}{R} \right)\)

\(\frac{1}{f} = (n – 1) \left( \frac{2}{R} \right)\)

Rearranging to solve for (R):

\(\frac{1}{R} = \frac{1}{2f(n – 1)}\)

\(R = 2f(n – 1)\)

Given Data

• Refractive index, (n = 1.55),
• Focal length, (f = 20 cm).

Substitute these values into the formula:

\(R = 2 \times 20 \times (1.55 – 1)\)

\(R = 40 \times 0.55\)

\(R = 22 \, \text{cm}\)

The radius of curvature required for each face of the double-convex lens is 22 cm.

Question 9.8: A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?

Solution 9.8: The object is virtual, and the image formed is real.

Object distance, u= +12 cm

(a) The focal length of the convex lens, f =20 cm

Image distance= v

\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\ \)

\(\frac{1}{v}-\frac{1}{12}=\frac{1}{20}\ \)

\(\frac{1}{v}=\frac{1}{20}+\frac{1}{12}\)

\(=\frac{3+5}{60}=\frac{8}{60}\ \)

Therefore v = \(\frac{60}{8}=7.5cm\)

The image will be formed 7.5cm away from the lens, to the right.

(b) Focal length of the concave lens, f = -16 cm

Image distance = v

\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\ \)

\(\frac{1}{v}=-\frac{1}{16}+\frac{1}{12}\)

\(=\frac{-3+4}{48}=\frac{1}{48}\)

∴ v= 48cm.

Hence, the image is formed 48 cm away from the lens, toward its right.

Question 9.9: An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

Solution 9.9: Size of the object, h₁ = 3cm

Object’s distance, u = -14 cm

The focal length of the concave lens, f = -21 cm

Image distance = v

According to the lens formula, we have the relation:

\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

\(\frac{1}{v}=-\frac{1}{21}-\frac{1}{14}\)

=\(\frac{-2-3}{42}=\frac{-5}{42}\)

Therefore v = \(-\frac{42}{5}=-8.4cm\)

Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.

The magnification of the image is given as:

\(m=\frac{Image\;height(h_{2})}{Object\;height(h_{1})}=\frac{v}{u}\)

Therefore h₂ =\(\frac{-8.4}{-14}\times3\)

\(=0.6\times3=1.8cm\)

The height of the image is 1.8 cm.

If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.

Question 9.10: What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

Solution 9.10: To find the effective focal length of a system consisting of a convex lens and a concave lens in contact, you can use the formula for the combined focal length of two lenses in contact:

\(\frac{1}{f_{\text{system}}} = \frac{1}{f_1} + \frac{1}{f_2}\)

where:

• ( f₁ ) is the focal length of the first lens (convex lens),
• ( f₂ ) is the focal length of the second lens (concave lens),
• ( f_system ) is the focal length of the combined lens system.

Given:

• Focal length of the convex lens ( f₁ ) = +30 cm (positive because it’s a convex lens),
• Focal length of the concave lens ( f₂ ) = -20 cm (negative because it’s a concave lens).

Plugging these values into the formula:

\(\frac{1}{f_{\text{system}}} = \frac{1}{30} + \frac{1}{-20}\)

Calculate each term:

\(\frac{1}{30} = 0.0333\)
\(\frac{1}{-20} = -0.05\)

Adding these:

\(\frac{1}{f_{\text{system}}} = 0.0333 – 0.05 = -0.0167\)

Thus:

f_system = \(\frac{1}{-0.0167} \approx -60 \text{ cm}\)

The focal length of the system is -60 cm.

Hence, the focal length of the combination of lenses is 60 cm. The negative sign indicates that the system of lenses acts as a diverging lens.

Question 9.11: A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Solution 9.11:

• The focal length of the objective lens, f₁ = 2.0 cm
• The focal length of the eyepiece, f₂ = 6.25 cm
• Distance between the objective lens and the eyepiece, d = 15 cm

(a) Least distance of distinct vision, d’ = 25 cm

Image distance for the eyepiece, V₂ = -25cm

Object distance for the eyepiece = u₂

According to the lens formula, we have the relation

\(\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}\)

\(\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}\)

= \(\frac{-5}{25}\)

Image distance for the objective lens,

\(v_{1}=d+u_{2}=15 – 5=10 cm\)

Object distance for the objective lens = \(u_{1}\)

\(\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}\)

\(\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}\)

\(\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}\)

\(u_{1}=-2.5cm\)

Magnifying power

\(m=\frac{v_{0}}{|u_{0}|}(1+\frac{d}{f_{e}}) \ \)

= \(\frac{10}{2.5}(1+\frac{25}{6.25}) \ \)

= 20

Hence, 20 is the magnifying power of the microscope.

(b) The final Image is formed at infinity.

Therefore, the image distance for the eyepiece, \(v_{2}=\infty\)

Object distance for the eyepiece = \( u_{2}\)

According to the lens formula, we have the relation

\(\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}\)

\(\frac{1}{\infty}-\frac{1}{u_{2}}=\frac{1}{6.25}\)

\(u_{2}=-6.25cm\)

\(u_{2}=-6.25cm\)

v₁ = d + u₂

=15 – 6.25

= 8.75 cm

The following relation is obtained from the lens formula:

\(\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}\)

\(\frac{1}{u_{1}}=\frac{1}{v_{1}}-\frac{1}{f_{1}}\)

\(\frac{1}{8.75}-\frac{1}{2.0}=\frac{2-8.75}{17.5}\)

\(u_{1}=-\frac{17.5}{6.75}\)

= -2.59 cm

Magnitude of the object distance,

\(|u_{1}|\) = 2.59 cm

Following is the relation explaining the magnifying power of a compound microscope:

m = \(\frac{v_{1}}{|u_{1}|}(\frac{d’}{|u_{2}|})\)

= \(\frac{8.75}{2.59}\times\frac{25}{6.25}\)

= 13.51

Hence, 13.51 is the magnifying power of the microscope.

Question 9.12: A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

Solution 9.12:

• The focal length of the objective lens, F_o = 8 mm = 0.8 cm
• The focal length of the eyepiece, F_e = 2.5 cm
• The object distance for the objective lens, u_o = -9.0 mm = -0.9 cm
• The least distance of distant vision, d = 25 cm
• The image distance for the eyepiece, v_e = -d = -25 cm
• The object distance for the eyepiece = u_e
• Using the lens formula, we can obtain the value of u_e as

\(\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}\)

\(\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}\)

\(\frac{1}{-25}-\frac{1}{2.5}=\frac{-1-10}{25}\)

\(u_{e}=-\frac{25}{11}\)

= -2.27cm

With the help of the lens formula, the value of the image distance for the objective (v) lens is obtained.

\(\frac{1}{v_{o}}-\frac{1}{u_{o}}=\frac{1}{f_{o}}\)

\(\frac{1}{v_{o}}=\frac{1}{f_{o}}+\frac{1}{u_{o}}\)

\(=\frac{1}{0.8}-\frac{1}{0.9}\)

\(=\frac{1}{7.2}\)

v_o = 7.2 cm

The separation between two lenses is determined as follows:

= v₀ + |u_e|

= 7.2 + 2.27

= 9.47 cm

The magnifying power of the microscope is calculated as follows:

Magnifying power, M = M_o × M_e

\(M = \frac{v_{o}}{u_{o}}(1+\frac{D}{f_{e}}) \)

\(M = \frac{7.2}{0.9}(1+\frac{25}{25})\)

= 8 × 11

= 88

Question 9.13: A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Solution 9.13: To solve this question, we need to calculate two things: the magnifying power of the telescope and the separation between the objective lens and the eyepiece.

1. Magnifying Power of the Telescope:

The magnifying power (M) of a telescope is given by the formula:

\(M = \frac{f_o}{f_e}\)

where:

• (f_o) is the focal length of the objective lens,
• (f_e) is the focal length of the eyepiece.

From the question:

• (f_o = 144 cm),
• (f_e = 6.0 cm).

Now, substituting the given values into the formula:

\(M = \frac{144}{6.0} = 24\)

So, the magnifying power of the telescope is 24.

2. Separation Between the Objective and the Eyepiece:

In a simple telescope focused on an object at infinity, the separation between the objective lens and the eyepiece is approximately the sum of their focal lengths. Therefore, the separation (d) is:

d =f_o + f_e

Substituting the values: d = 144 cm + 6.0 cm = 150 cm

So, the separation between the objective and the eyepiece is 150 cm.

• The magnifying power of the telescope is 24.
• The separation between the objective and the eyepiece is 150 cm.

Question 9.14: (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope? (b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 10⁶m, and the radius of lunar orbit is 3.8 × 10⁸m.

Solution 9.14:

The focal length of the objective lens, f_o = 15m = 15 × 10²cm

The focal length of the eyepiece, f_e = 1.0 cm

(a) \(\alpha =\frac{f_{o}}{f_{e}}\) is the angular magnification of a telescope.

= \(\frac{15\times10^{2}}{1}\)

= 1500

Hence, the angular magnification of the given refracting telescope is 1500.

(ii) Diameter of the moon, d = 3.48 × 10⁶m

Radius of the lunar orbit, r_o = 3.8 × 10⁸m

Consider d’ to be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter of the moon is equal to the angle subtended by
the image.

\(\frac{d}{r_{o}}=\frac{d’}{f_{o}}\)

\(\frac{3.48\times10^{6}}{3.8\times10^{8}}=\frac{d’}{15}\)

Therefore, d’=

\(\frac{3.48}{3.8}\times10^{-2}\times15\)

\(13.74\times10^{-2}m\)

= 13.74cm

Therefore, 13.74cm is the diameter of the image of the moon formed by the objective lens.

Question 9.15: Use the mirror equation to deduce that:
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]

Solution 9.15:

(i) The focal length is negative for a concave mirror.
Therefore, f < 0
The object distance (u) is negative when the object is placed on the left side of the mirror.
Therefore, u < 0
Lens formula for image distance v is written as

\(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)

\(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)

(a) For a concave mirror, f is negative, i.e., f < 0

For a real object, i.e., which is on the left side of the mirror.

For u between f and 2f implies that 1/u lies between 1/f and 1/2f

\(\frac{1}{2f}>\frac{1}{u}>\frac{1}{f}(as \, u, \, f \, are \, negative) \)

\(\frac{1}{f}-\frac{1}{2f}<\frac{1}{f}-\frac{1}{u}<0 \ \)

\(\frac{1}{2f}<\frac{1}{v}<0\)

1/v is negative

which implies that v is negative and greater than 2f. Hence, the image lies beyond 2f, and it is real.

(b) The focal length is positive for a convex mirror.
Therefore, f > 0
The object distance (u) will be negative if the object is placed on the left side of the mirror.
Therefore, u < 0
Using the mirror formula, we can calculate the image distance v.

\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

\(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)

Using equation (ii), we can conclude that \(\frac{1}{v}<0\)

v>0

Thus, the image is obtained on the back side of the mirror. Therefore, it can be concluded that a convex mirror always produces a virtual image irrespective of the object’s distance.

(c) The focal length is positive for a convex mirror.
Therefore, f > 0
The object distance (u) is negative when the object is placed on the left side of the mirror
Therefore, u < 0
For image distance v, we have the mirror formula.

\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

\(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)

But we have u<0

Therefore, \(\frac{1}{v}>\frac{1}{f}\)

v<f

Therefore, the image is formed between the focus and the pole and is diminished.

(d) The focal length of the concave mirror is negative.

Therefore, f < 0

The object distance, u, is negative for an object that is placed on the left side of the mirror.
Therefore, u < 0
It is placed between the focus (f) and the pole.

Therefore, f>u>0

\(\frac{1}{f}<\frac{1}{u}<0\)

\(\frac{1}{f}-\frac{1}{u}<0\)

For image distance v, we have the mirror formula.

\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

\(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)

Therefore, \(\frac{1}{v}<0\)

v>0

The image obtained is virtual since it is formed on the right side of the mirror.
For u < 0 and v > 0, we can write \(\frac{1}{u}>\frac{1}{v}\)

Magnification, m = \(\frac{v}{u}\) > 1

Hence, the formed image is enlarged.

Question 9.16: A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?

Solution 9.16:

The actual depth of the pin, d = 15 cm
The apparent depth of the pin = d’
The refractive index of glass, μ = 1.5

The ratio of actual depth to the apparent depth and the refractive index of the glass are equal.

i.e., \(\mu=\frac{d}{d’}\)

Therefore, d’= \(\frac{d}{\mu}\)

= \(\frac{15}{1.5}\)

= 10cm

The distance at which the pin appears to be raised =d’-d=15-10=5cm When the angle of incidence is small, the distance is independent of the location of the slab.

Question 9.17: (a) Figure 9.28 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure. (b) What is the answer if there is no outer covering of the pipe?

Solution 9.17:

(i) Refractive index of the glass-fibre, μ₂ = 1.68

The Refractive index of the outer covering, μ₁ = 1.44

The angle of incidence = i
The angle of refraction = r
The angle of incidence at the interface = i’
The refractive index ( μ ) of the inner core-outer core interface is given as

\(\mu=\frac{\mu_{2}}{\mu_{1}}=\frac{1}{sin i’}\)

\(sin i’=\frac{\mu_{1}}{\mu_{2}}\)

\(=\frac{1.44}{1.68}=0.8571\)

\(Therefore\;i’=59°\)

For the critical angle, total internal reflection (TIR) takes place only when i>i’, i.e., i> 59°

Maximum angle of reflection,

\(r_{max}=90°-i’=90°-59°=31°\)

Let i_maz be the maximum angle of incidence.

The refractive index at the air-glass interface, μ₁ = 1.68

We know that \(\mu_{1}=\frac{sin i_{max}}{sin r_{max}}\) Rearranging the equation, we get sin i_max = 1.68 × sin r_max

Substituting the values in the above equation, we get

= 1.68 sin 31°

= 1.68 × 0.5150

= 0.8652

Therefore,

\(i_{max}=sin^{-1}0.8652=60°\)

Thus, all the rays incident at angles lying in the range 0 < i < 60°

will suffer total internal reflection.

(b) If the outer covering of the pipe is not present, then Refractive index of the outer pipe, μ₁ =Refractive index of air=1

For the angle of incidence i = 90°,

We can write Snell’s law at the air-pipe interface as

\(\frac{sin i}{sin r}=\mu_{2}=1.68\)

\(sin r=\frac{sin 90°}{1.68}=\frac{1}{1.68}\)

\(r=sin^{-1}(0.5932)\)

= 36.5°

Therefore,

\(i’=90°-36.5°=53.5°\)

Since i’>r, all incident rays will suffer total internal reflection.

Question 9.18: The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

Solution 9.18:

Distance between the object and the image, d = 3 m

Maximum focal length of the convex lens = F_maz

For real images, the maximum focal length is given as \(f_{max}=\frac{d}{4}\)

= \(\frac{3}{4}=0.75m\)

Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.

Question 9.19: A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.

Solution 9.19:

Distance between the image (screen) and the object, D = 90 cm
Distance between two locations of the convex lens, d = 20 cm
The focal length of the lens = f
The focal length is related to d and D as \(f=\frac{D^{2}-d^{2}}{4D}\)

= \(\frac{90^{2}-(20^{2})}{4\times90}\)

= \(\frac{770}{36}=21.39cm\)

Therefore, the focal length of the convex lens is 21.39 cm.

Question 9.20: (a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

Solution 9.20:

The focal length of the convex lens, F₁ = 30 cm

The focal length of the concave lens, F₂ = -20 cm

The distance between the two lenses, d = 8.0 cm

(a) When the parallel beam of light is incident on the convex lens first,
Using the lens formula, we can write

\(\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}\)

Where,

u₁ = Image distance

\(\frac{1}{v_{1}}=\frac{1}{30}-\frac{1}{\infty}=\frac{1}{30}\)

Therefore, \(v_{1}=30cm\)

For a concave lens, the image acts as a virtual object.
Using the lens formula for the concave lens, we can write

\(\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}\)

Where,

u₂ = Object distance

= (30 − d) = 30 − 8 = 22 cm

V₂ = Image distance

\(\frac{1}{v_{2}}=\frac{1}{22}-\frac{1}{20}\)

\(= \frac{10-11}{220}=\frac{-1}{220}\)

Therefore, V₂ = -220cm

 The parallel incident beam appears to diverge from a point that is 

\((220-\frac{d}{2}=220-4)216cm\)

from the centre of the combination of the two lenses.
When the parallel beam of light is incident, from the left, on the concave lens
first,
Using the lens formula, we can write

\(\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}\)

\(\frac{1}{v_{2}}=\frac{1}{f_{2}}+\frac{1}{u_{2}}\)

Where,

u₂ = Object distance = −∞

V₂ = Image distance

\(\frac{1}{v_{2}}=\frac{1}{-20}+\frac{1}{-\infty}=-\frac{1}{20}\)

Therefore, V₂ = -20cm

For a convex lens, the image will act as a real object.
Applying the lens formula to the convex lens, we have

\(\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}\)

Where,

u₁ = Object distance

= −(20 + d) = −(20 + 8) = −28 cm

V₁ = Image distance

\(\frac{1}{v_{1}}=\frac{1}{30}+\frac{1}{-28}\)

\( = \frac{14-15}{420}=\frac{-1}{420}\)

Therefore, V₂ = -420cm

The parallel incident beam appears to diverge from (420-4) 416cm. The diversion happens from the left of the centre of the combination of the two lenses. The answer is dependent on the side of the combination where the incident beam of light is parallel.

(b) Height of the image,

h₁ = 1.5 cm

Object distance from the side of the convex lens, u₁ = -40 cm

\(|u_{1}|=40cm\)

Using the lens formula,

\(\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}\)

Where,

v₁ = Image distance

\(\frac{1}{v_{1}}=\frac{1}{30}+\frac{1}{-40}\)

\(= \frac{4-3}{120}=\frac{1}{120}\)

Therefore, v₁ = 120cm

Magnification, m = \(\frac{v_{1}}{|u_{1}|}=\frac{120}{40}=3\)

Therefore, 3 is the magnification of the convex lens.
The image of the convex lens is an object for the concave lens.
According to the lens formula,

\(\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}\)

Where,

u₂ = Object distance

= +(120 − 8) = 112 cm.

V₂ = Image distance

\(\frac{1}{v_{2}}=\frac{1}{-20}+\frac{1}{112}\)

\(= \frac{-112+20}{2240}=\frac{-92}{2240}\)

Therefore \(v_{2}=\frac{-2240}{92}\)

Magnification, m’= \(|\frac{v_{2}}{u_{2}}|=\frac{2240}{92}\times \frac{1}{112}=\frac{20}{92}\)

Hence, the magnification due to the concave lens is \(\frac{20}{92}\)

The magnification produced by the combination of the two lenses is calculated as \(m \times m’\)

= 3 × 20/92 = 60/92 = 0.652

h₂ = h₁ × 0.652

Where,

h₁ = Object size = 1.5 cm

h₂ = Size of the image

Therefore
\(h_{2}=0.652\times1.5=0.98cm\)

Hence, the height of the image is 0.98 cm.