Chapter 10 Wave Optics NCERT Solutions Class 12 Physics

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.

Given Data

$\lambda_{air} = 589 \text{ nm} = 589 \times 10^{-9} \text{ m}$
Speed of light in air, $c = 3 \times 10^8 \text{ m/s}$
Refractive index of water, $\mu = 1.33$

(a) Reflected Light

Reflection occurs in the same medium (air). Therefore, speed, wavelength, and frequency remain unchanged.

Speed: $v = c = 3 \times 10^8 \text{ m/s}$
Wavelength: $\lambda = 589 \text{ nm}$
Frequency: $\nu = c / \lambda$

$$\nu = \frac{3 \times 10^8}{589 \times 10^{-9}} = 5.09 \times 10^{14} \text{ Hz}$$

(b) Refracted Light

Refraction occurs in water. Frequency remains constant (source property), but speed and wavelength change.

Frequency: Same as air, $\nu = 5.09 \times 10^{14} \text{ Hz}$
Speed: $v = c / \mu$

$$v = \frac{3 \times 10^8}{1.33} \approx 2.26 \times 10^8 \text{ m/s}$$

Wavelength: $\lambda_{water} = \lambda_{air} / \mu$

$$\lambda_{water} = \frac{589}{1.33} \approx 444 \text{ nm}$$

(a) Reflected: $\lambda=589 \text{ nm}, \nu=5.09 \times 10^{14} \text{ Hz}, v=3 \times 10^8 \text{ m/s}$
(b) Refracted: $\lambda=444 \text{ nm}, \nu=5.09 \times 10^{14} \text{ Hz}, v=2.26 \times 10^8 \text{ m/s}$

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.

Solutions

(a) Spherical Wavefront: Light spreads out in all directions from a point source, forming spheres.
(b) Plane Wavefront: Rays emerging from the focus become parallel after passing through a convex lens. Parallel rays correspond to a plane wavefront.
(c) Plane Wavefront: A star is effectively a point source at infinity. A small portion of a huge spherical wavefront intercepted at a large distance appears flat (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 $3.0 \times 10^8 \text{ m s}^{-1}$)
(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?

(a) Speed Calculation

$$v = \frac{c}{\mu} = \frac{3 \times 10^8}{1.5} = 2.0 \times 10^8 \text{ m/s}$$

(b) Dependence on Colour

No, the speed of light in glass depends on the colour (wavelength). This phenomenon is called dispersion.

Refractive index $\mu$ increases as wavelength $\lambda$ decreases (Cauchy’s relation). Since $\lambda_{violet} < \lambda_{red}$, we have $\mu_{violet} > \mu_{red}$.

Since $v = c/\mu$, a higher $\mu$ means a lower speed.

Violet light travels slower than red light in glass.

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.

[Image of Young’s Double Slit Experiment Diagram]

Given Data

Slit separation, $d = 0.28 \text{ mm} = 0.28 \times 10^{-3} \text{ m}$
Screen distance, $D = 1.4 \text{ m}$
Position of 4th bright fringe, $x_4 = 1.2 \text{ cm} = 1.2 \times 10^{-2} \text{ m}$
Order of fringe, $n = 4$

Formula & Calculation

For bright fringes: $x_n = \frac{n \lambda D}{d}$

$$\lambda = \frac{x_n d}{n D} = \frac{(1.2 \times 10^{-2}) \times (0.28 \times 10^{-3})}{4 \times 1.4}$$ $$\lambda = \frac{0.336 \times 10^{-5}}{5.6} = 0.06 \times 10^{-5} \text{ m}$$ $$\lambda = 600 \times 10^{-9} \text{ m} = 600 \text{ nm}$$

Wavelength $\lambda = 600 \text{ nm}$.

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 $\lambda/3$?

Formula

Intensity $I$ at a point with phase difference $\phi$ is given by $I = I_0 \cos^2(\phi/2)$, where $I_0$ is the maximum intensity.

Relation between Phase Diff ($\phi$) and Path Diff ($\Delta x$): $\phi = \frac{2\pi}{\lambda} \Delta x$.

Case 1: Path Diff = $\lambda$

$$\phi_1 = \frac{2\pi}{\lambda} (\lambda) = 2\pi$$ $$K = I_0 \cos^2\left(\frac{2\pi}{2}\right) = I_0 \cos^2(\pi) = I_0 (-1)^2 = I_0$$

So, Maximum Intensity $I_0 = K$.

Case 2: Path Diff = $\lambda/3$

$$\phi_2 = \frac{2\pi}{\lambda} \left(\frac{\lambda}{3}\right) = \frac{2\pi}{3} = 120^\circ$$ $$I = K \cos^2\left(\frac{120^\circ}{2}\right) = K \cos^2(60^\circ)$$ $$I = K \left(\frac{1}{2}\right)^2 = \frac{K}{4}$$

Intensity = 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?

Given Data

$\lambda_1 = 650 \text{ nm}$, $\lambda_2 = 520 \text{ nm}$.
Assume standard values if not given (though usually $D$ and $d$ are needed for absolute distance). Note: The question implies finding expression or value if D/d were given. Since D and d are not provided in problem statement text, we usually solve in terms of D/d or assume the values from Q10.4 if linked (D=1.2m, d=2mm are common in NCERT examples, but let’s solve generally or check if D,d are implicit. Wait, usually this question is standalone. We will calculate x in terms of D/d or assuming standard D=1.2m, d=2mm is not safe. Let’s calculate $x$ as a formula or assume data is missing in the prompt. Ah, checking NCERT text, usually D=1.2m and d=2mm are given in similar solved examples, but strictly speaking, without D and d, we can only find ratio or ‘least distance’ as a multiple of D/d. However, often in exams D=1.2m, d=2mm is used. Let’s provide the general formula and the condition for coincidence.)

Assumption for calculation: Let’s calculate the condition for (b) and formula for (a).

(a) 3rd Bright Fringe for $\lambda_1$

$$x = \frac{n \lambda_1 D}{d} = \frac{3 \times 650 \times 10^{-9} D}{d} = 1950 \frac{D}{d} \text{ nm}$$

(b) Coincidence Condition

Fringes coincide when $x = x_1 = x_2$. Let $n_1$ th fringe of $\lambda_1$ coincide with $n_2$ th fringe of $\lambda_2$.

$$\frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d} \implies n_1 \lambda_1 = n_2 \lambda_2$$ $$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{520}{650} = \frac{4}{5}$$

Least integral values are $n_1 = 4$ and $n_2 = 5$.
So, the 4th bright fringe of 650 nm coincides with the 5th bright fringe of 520 nm.

$$x_{min} = \frac{4 \lambda_1 D}{d} = \frac{4 \times 650 \times 10^{-9} D}{d} = 2600 \frac{D}{d} \text{ nm}$$

(a) $x_3 = 1950 (D/d) \text{ nm}$
(b) Coincidence at $n_1=4$ ($\lambda_1$) and $n_2=5$ ($\lambda_2$). Distance $x = 2600 (D/d) \text{ nm}$.