## Wave Optics ncert solutions: Ncert Class 12 Physics Chapter 10 Exercise Solutions

Textbook | NCERT |

Class | Class 12 |

Subject | Physics |

Chapter | Chapter 10 |

Chapter Name | Wave Optics class 12 exercise solutions |

Category | Ncert Solutions |

Medium | English |

Are you looking for Ncert Solution for Class 12 Physics Chapter 10 Wave Optics? Now you can download Ncert Class 12 Physics Chapter 10 Exercise Solutions pdf from here.

### Question 10.1: Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of

(a) reflected, and (b) refracted light? Refractive index of water is 1.33.

**Solution 10.1:**

Monochromatic light incident having wavelength,

\(\lambda\) = 589 nm = 589 x 10^{-9} m

Speed of light in air, c = 3 x 10^{8} m s^{-1}

Refractive index of water, μ = 1.33

(a) The ray will be reflected back to the same medium through which it passed.

Therefore, the wavelength, speed and frequency of the reflected ray will be the same as that of the incident ray.

The frequency of light can be found from the relation: \(v=\frac{c}{\lambda}\)

\(\frac{3\times10^{8}}{589\times10^{-9}}\)

= 5.09 × 10^{14} Hz

Hence, **c = 3 x 10 ^{8} m s^{-1}, 5.09 × 10^{14} Hz**, and 589 nm

**are the speed, frequency and wavelength of the reflected light.**

b) The frequency of light which is travelling never depends upon the property of the medium. Therefore, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in the air.

Refracted frequency, v = 5.09 x 10^{14} Hz

Following is the relation between the speed of light in water and the refractive index of the water: \(v=\frac{c}{\mu}\)

= \(v=\frac{3\times10^{8}}{1.33}\)

= 2.26 × 10^{8} m s^{‑1}

Below is the relation for finding the wavelength of light in water: \(\lambda=\frac{v}{V}\)

\(\frac{2.26\times10^{8}}{5.09\times10^{14}}\)

= 444.007 × 10^{-9 }m** **= 444.01 nm

**Therefore, 444.007 × 10 ^{-9} m, 444.01nm, and 5.09 × 10^{14} Hz **are the speed, frequency, and wavelength of the refracted light.

### Question 10.2: What is the shape of the wavefront in each of the following cases:

(a) Light diverging from a point source.

(b) Light emerging out of a convex lens when a point source is placed at its focus.

(c) The portion of the wavefront of light from a distant star intercepted by the Earth.

**Solution 10.2:** (a) The shape of the wavefront in case of a light diverging from a point source is spherical. The wavefront emanating from a point source is shown in the given figure.

(b) The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid. This is shown in the given figure.

(c) The portion of the wavefront of light from a distant star intercepted by the Earth is a plane.

### Question 10.3: (a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is 3.0 **×** 10^{8} m s^{–1})

(b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?

**Solution 10.3:** **(a)** Refractive Index of glass, μ = 1.5

Speed of light, c = 3 × 10^{8} ms^{-1}

The relation for the speed of light in a glass is: \(v=\frac{c}{\mu}\)

= \(\frac{3\times10^{8}}{1.5}\)

= 2 × 10^{8} m s^{-1}

**Hence, the speed of light in glass is 2 × 10 ^{8} m s^{-1}**

**(b)** The speed of light is dependent on the colour of the light. For a white light, the refractive index of the violet component is greater than the refractive index of the red component. So, the speed of violet light is less than the speed of red light in the glass. This will reduce the speed of violet light in a glass prism when compared with red light.

### Question 10.4: In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

**Solution 10.4:** To determine the wavelength of light used in the Young’s double-slit experiment, we can apply the formula for the position of bright fringes. The distance between bright fringes in a double-slit experiment is given by:

\(y_m = \frac{m \lambda L}{d}\)

Where:

- ( y
_{m}) is the distance between the central bright fringe and the m -th bright fringe. - ( m ) is the order of the bright fringe (for the fourth bright fringe, ( m = 4 )).
- \(( \lambda )\) is the wavelength of the light.
- ( L ) is the distance from the slits to the screen.
- ( d ) is the separation between the slits.

**Given values:**

- d = 0.28 mm = 0.28
**×**10^{-3}m - L = 1.4 m
- Distance between the central bright fringe and the fourth bright fringe is y
_{4}= 1.2 cm = 1.2**×**10^{-2}m - ( m = 4 ) (since it’s the fourth bright fringe)

**Rearranging the equation to solve for the wavelength \( \lambda \):**

\(\lambda = \frac{y_m d}{m L}\)

Substituting the known values:

\(\lambda = \frac{(1.2 \times 10^{-2} \, \text{m}) \times (0.28 \times 10^{-3} \, \text{m})}{4 \times 1.4 \, \text{m}}\)

Now, let’s calculate the wavelength.

\(\lambda = \frac{(1.2 \times 10^{-2}) \times (0.28 \times 10^{-3})}{4 \times 1.4}\)

\(\lambda = \frac{3.36 \times 10^{-6}}{5.6}\)

\(\lambda = 6 \times 10^{-7} \, \text{m} = 600 \, \text{nm}\)

The wavelength of the light used in the experiment is \( \lambda = 600 \, \text{nm} \).

### Question 10.5: In Young’s double-slit experiment using monochromatic light of wavelength \( \lambda \), the intensity of light at a point on the screen where path difference is \( \lambda \), is K units. What is the intensity of light at a point where path difference is \( \lambda \)/3?

**Solution 10.5:** In a Young’s double-slit experiment, the intensity of light at any point on the screen depends on the phase difference between the light waves from the two slits. This phase difference is related to the path difference.

**Formula for Intensity:**

The intensity ( I ) at a point on the screen is related to the phase difference \( \Delta \phi \) between the two waves as:

\(I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right)\)

Where:

- I
_{0}is the maximum intensity (which occurs when the phase difference is zero, i.e., the waves are in phase). - \( \Delta \phi \) is the phase difference.

The phase difference\( \Delta \phi \) is related to the path difference \( \Delta x \) by:

\(\Delta \phi = \frac{2\pi}{\lambda} \Delta x\)

Where \( \lambda \) is the wavelength of the light.

**Case 1: Path Difference = \( \lambda \)**

When the path difference is \( \Delta x = \lambda \), the phase difference is:

\(\Delta \phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi\)

At this point, the intensity is given as ( K ). The intensity at a phase difference of \( 2\pi \) is:

\(I = I_0 \cos^2\left(\frac{2\pi}{2}\right)\)

\(= I_0 \cos^2(\pi) \)

\(= I_0 \times 0 = 0\)

However, since the problem states that the intensity is ( K ), we interpret this to mean ( K ) is the intensity relative to the maximum, which occurs when the waves are exactly in phase. This likely implies a different baseline (not zero intensity) for this particular point.

Case 2: Path Difference = \( \lambda/3 \)

Now, when the path difference is \( \Delta x = \lambda/3 \), the phase difference is:

\(\Delta \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3}\)

The intensity at this phase difference is:

\(I = I_0 \cos^2\left(\frac{2\pi}{3} \times \frac{1}{2}\right)\)

\(= I_0 \cos^2\left(\frac{\pi}{3}\right)\)

Using \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \), we get:

\(I = I_0 \times \left(\frac{1}{2}\right)^2 = \frac{I_0}{4}\)

Thus, the intensity at a point where the path difference is \( \lambda/3 \) is \( \frac{1}{4} \) of the maximum intensity I_{0}.

Since the intensity at \( \Delta x = \lambda \) is given as ( K ), and using ( K = I_{0} ), we have:

\(I = \frac{K}{4}\)

The intensity of light at a point where the path difference is \( \lambda/3 \) is \( \frac{K}{4}\) units.

### Question 10.6: A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.

(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.

(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

**Solution 10.6:**

Wavelength of the light beam, \(\lambda_{1}\) **= 650 nm**

Wavelength of another light beam, \(\lambda_{2}\) **= 520 nm**

Distance of the slits from the screen = D

Distance between the two slits = d

**(a)** Distance of the n^{th} bright fringe on the screen from the central maximum is given by the relation,

x = \(n\;\lambda_{1}\;(\frac{D}{d})\)

For the third bright fringe, n = 3

Therefore, x = \(3\times650\frac{D}{d}= 1950\frac{D}{d}nm\)

**(b)** Let, the n^{th} bright fringe due to wavelength \(\lambda_{2}\) and ( n-1)^{th} bright fringe due to wavelength \(\lambda_{2}\) coincide on the screen. The value of n can be obtained by equating the conditions for bright fringes:

\(n\lambda_{2}=(n-1)\lambda_{1}\)

520n = 650n – 650

650 = 130n

Therefore, n = 5

Hence, the least distance from the central maximum can be obtained by the relation:

x = \(n\;\lambda_{2}\;\frac{D}{d}\)

= \(5\times 520\frac{D}{d}=2600\frac{D}{d}\) nm

Note: The value of d and D are not given in the question.

### Question 10.7: In a double-slit experiment, 0.2° is found to be the angular width of a fringe on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take the refractive index of water to be \(\frac{4}{3}\).

**Solution 10.7:** In a double-slit experiment, the angular width of a fringe is related to the wavelength of light and the slit separation. The angular width \( \theta \) of the fringe is given by:

\(\theta = \frac{\lambda}{d}\)

where:

- \( \lambda \) is the wavelength of light,
- d is the separation between the slits.

**Part 1: Angular Width in Air**

In the given problem, the angular width in air is 0.2° and the wavelength of light used is \( (\lambda = 600 \, \text{nm} ) (or ( 600 \times 10^{-9} \, \text{m} )\).

Let’s convert 0.2° to radians:

\(0.2^\circ = 0.2 \times \frac{\pi}{180} \, \text{radians} \)

\(\approx 3.49 \times 10^{-3} \, \text{radians}\)

Thus, the angular width in air is:

\(\theta_{\text{air}} = 3.49 \times 10^{-3} \, \text{radians}\)

Since \(\theta = \frac{\lambda}{d}\), we can solve for ( d ) (the slit separation) using the wavelength in air:

\(d = \frac{\lambda_{\text{air}}}{\theta_{\text{air}}} \)

\(= \frac{600 \times 10^{-9}}{3.49 \times 10^{-3}} \)

\(\approx 1.72 \times 10^{-4} \, \text{m}\)

**Part 2: Angular Width in Water**

When the entire apparatus is immersed in water, the wavelength of light decreases because the wavelength in a medium is given by:

\(\lambda_{\text{water}} = \frac{\lambda_{\text{air}}}{n}\)

where ( n ) is the refractive index of water \(( n = \frac{4}{3} )\).

Thus, the wavelength in water is:

\(\lambda_{\text{water}} = \frac{600 \times 10^{-9}}{4/3} = 450 \times 10^{-9} \, \text{m}\)

The angular width of the fringe in water is:

\(\theta_{\text{water}} = \frac{\lambda_{\text{water}}}{d}\)

Substituting the values:

\(\theta_{\text{water}} = \frac{450 \times 10^{-9}}{1.72 \times 10^{-4}} \)

\(\approx 2.62 \times 10^{-3} \, \text{radians}\)

Now, convert this result back to degrees:

\(\theta_{\text{water}} = 2.62 \times 10^{-3} \times \frac{180}{\pi} \approx 0.15^\circ\)

The angular width of the fringe when the apparatus is immersed in water is 0.15°.

### Question 10.8: What is the Brewster angle for air-to-glass transition? (Refractive index of glass = 1.5.)

**Solution 10.8:** The Brewster angle, \( \theta_B \), is the angle of incidence at which light with a particular polarization is perfectly transmitted through a surface without any reflection. It occurs when the reflected and refracted light rays are perpendicular to each other. The Brewster angle is given by the formula:

\(\tan \theta_B = \frac{n_2}{n_1}\)

Where:

- \( \theta_B\) is the Brewster angle,
- n
_{2}is the refractive index of the second medium (glass, in this case), - n
_{1}is the refractive index of the first medium (air).

For air, n_{1} = 1, and for glass, n_{2} = 1.5. Using the formula:

\(\tan \theta_B = \frac{1.5}{1} = 1.5\)

Now, we calculate \( \theta_B \) by taking the arctangent (inverse tangent) of 1.5:

\(\theta_B = \tan^{-1}(1.5)\)

Using a calculator:

\(\theta_B \approx 56.31^\circ\)

The Brewster angle for air-to-glass transition is approximately 56.3°.

### Question 10.9: Light of wavelength 5000 Armstrong falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?

**Solution 10.9:** **Part 1: Wavelength and Frequency of Reflected Light**

When light is reflected from a plane surface, **neither the wavelength nor the frequency of the light changes**. The frequency of light depends only on the source, and the wavelength in air (or vacuum) remains the same as long as the light continues to travel in the same medium.

**Given Data:**

- Wavelength of incident light \( \lambda = 5000 \, \text{Å} \)
- Speed of light \( c = 3 \times 10^8 \, \text{m/s} \)
- 1 Ångström \( = 10^{-10} \, \text{m} \)

**Wavelength of the reflected light:**

Since the medium does not change, the wavelength of the reflected light remains the same as the incident light:

\(\lambda_{\text{reflected}} = \lambda_{\text{incident}} \)

\(= 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \)

\(= 5 \times 10^{-7} \, \text{m}\)

**Frequency of the reflected light:**

The frequency of light is given by the equation:

\(f = \frac{c}{\lambda}\)

Substituting the known values:

\(f = \frac{3 \times 10^8 \, \text{m/s}}{5 \times 10^{-7} \, \text{m}} \)

\(= 6 \times 10^{14} \, \text{Hz}\)

Thus, the **wavelength** of the reflected light is \( 5000 \, \text{Å} \), and the **frequency** is \( 6 \times 10^{14} \, \text{Hz} \).

**Part 2: Angle of Incidence for Reflected Ray to Be Normal to Incident Ray**

For the reflected ray to be normal (perpendicular) to the incident ray, the angle between the incident ray and the reflected ray must be \( 90^\circ \). In a reflection, the angle of incidence \( \theta_i \) is equal to the angle of reflection \( \theta_r \).

The angle between the incident and reflected rays is \( 2 \theta_i \). For this to be \( 90^\circ \), we must have:

\(2 \theta_i = 90^\circ\)

Thus:

\(\theta_i = \frac{90^\circ}{2} = 45^\circ\)

**Wavelength**of reflected light: \( 5000 \, \text{Å} \)**Frequency**of reflected light: \(6 \times 10^{14} \, \text{Hz} \)- The
**angle of incidence**for the reflected ray to be normal to the incident ray is \( 45^\circ \).

### Question 10.10: Estimate the distance for which ray optics is a good approximation for an aperture of 4 mm and wavelength of 400 nm.

**Solution 10.10:** Fresnel’s distance ( Z_{F} ) is the distance which is used in ray optics for a good approximation. Following is the relation, \(Z_{F}=\frac{a^{2}}{\lambda}\)

Where,

Aperture width, a = 4 mm = 4 × 10^{-3} m

Wavelength of light,

\(\lambda\) = 400 nm = 400 × 10^{-9} m

\(Z_{F}=\frac{(4\times10^{-3})^{2}}{400\times10^{-9}}\)

= 40m

Therefore, 40 m is the distance for which the ray optics is a good approximation.