Q: Assume that, light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2 m is

3.66 \times 10^{-7} rad. Limit of resolution of a telescope: d\theta = \frac{1.22\lambda}{d} = \frac{1.22 \times 600 \times 10^{-9}}{2} = 3.66 \times 10^{-7} rad

Q: Angular width of the central maxima in the Fraunhoffer diffraction for \lambda = 6000 Å is \theta_0. When the same slit is illuminated by another monochromatic light, the angular width decreases by 30%. The wavelength of this light is

4200 Å. Angular width of central maximum: 2\theta = \frac{2\lambda}{a}, so \theta \propto \lambda (slit width constant). New angular width = 0.7\theta_0, so \lambda_2 = 0.7 \times \lambda_1 = 0.7 \times 6000 = 4200 Å

Q: In a diffraction pattern due to a single slit of width a, the first minimum is observed at an angle 30° when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of

\sin^{-1}\left(\frac{3}{4}\right). First minimum: a\sin 30° = \lambda \Rightarrow a = 2\lambda First secondary maximum: a\sin\theta_1 = \frac{3\lambda}{2} \sin\theta_1 = \frac{3\lambda}{2a} = \frac{3\lambda}{4\lambda} = \frac{3}{4} \theta_1 = \sin^{-1}\left(\frac{3}{4}\right)

Q: For a parallel beam of monochromatic light of wavelength \lambda, diffraction is produced by a single slit whose width a is of the order of the wavelength of the light. If D is the distance of the screen from the slit, the width of the central maxima will be

\frac{2D\lambda}{a}. For the first minimum, \sin\theta = \frac{\lambda}{a}, and for small angles \sin\theta = \frac{Y}{D} Y = \frac{\lambda D}{a} Width of central maximum = 2Y = \frac{2\lambda D}{a}

Question 1

Assume that, light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2 m is

Accepted Answer

3.66 	imes 10^{-7} rad. Limit of resolution of a telescope:

d	heta = \frac{1.22\lambda}{d} = \frac{1.22 	imes 600 	imes 10^{-9}}{2} = 3.66 	imes 10^{-7} rad

Question 2

The Brewster's angle i_b for an interface should be

Accepted Answer

45° < i_b < 90°. The Brewster's angle for an interface satisfies 	an i_b = \mu. Since \mu > 1 for any denser medium, i_b lies between 45° and 90°.

Question 3

Angular width of the central maxima in the Fraunhoffer diffraction for \lambda = 6000 Å is 	heta_0. When the same slit is illuminated by another monochromatic light, the angular width decreases by 30%. The wavelength of this light is

Accepted Answer

4200 Å. Angular width of central maximum: 2	heta = \frac{2\lambda}{a}, so 	heta \propto \lambda (slit width constant).

New angular width = 0.7	heta_0, so

\lambda_2 = 0.7 	imes \lambda_1 = 0.7 	imes 6000 = 4200 Å

Question 4

An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of

Accepted Answer

large focal length and large diameter. Angular magnification M = \frac{f_0}{f_e} — for large magnification the objective should have a large focal length f_0.

Angular resolution R = \frac{a}{1.22\lambda} — for high resolution the objective should have a large diameter a.

Hence the objective lens should have a large focal length and a large diameter.

Question 5

Unpolarised light is incident from air on a plane surface of a material of refractive index \mu. At a particular angle of incidence i, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?

Accepted Answer

Reflected light is polarised with its electric vector perpendicular to the plane of incidence. When the reflected and refracted rays are perpendicular to each other, the angle of incidence is the Brewster's angle (	an i = \mu, not 	an i = 1/\mu). At this angle the reflected light is completely plane polarised, with its electric vector perpendicular to the plane of incidence.

Question 6

The ratio of resolving powers of an optical microscope for two wavelengths \lambda_1 = 4000 Å and \lambda_2 = 6000 Å is

Accepted Answer

3 : 2. Resolving power of a microscope \propto \frac{1}{\lambda}

\frac{RP_1}{RP_2} = \frac{\lambda_2}{\lambda_1} = \frac{6000}{4000} = \frac{3}{2}

Question 7

Two polaroids P_1 and P_2 are placed with their axis perpendicular to each other. Unpolarised light I_0 is incident on P_1. A third polaroid P_3 is kept in between P_1 and P_2 such that its axis makes an angle 45° with that of P_1. The intensity of transmitted light through P_2 is

Accepted Answer

\frac{I_0}{8}. After P_1: I_1 = \frac{I_0}{2}

After P_3 (at 45° to P_1): I_2 = \frac{I_0}{2}\cos^2 45° = \frac{I_0}{4}

After P_2 (at 45° to P_3): I_3 = \frac{I_0}{4}\cos^2 45° = \frac{I_0}{8}

Question 8

In a diffraction pattern due to a single slit of width a, the first minimum is observed at an angle 30° when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of

Accepted Answer

\sin^{-1}\left(\frac{3}{4}ight). First minimum: a\sin 30° = \lambda \Rightarrow a = 2\lambda

First secondary maximum: a\sin	heta_1 = \frac{3\lambda}{2}

\sin	heta_1 = \frac{3\lambda}{2a} = \frac{3\lambda}{4\lambda} = \frac{3}{4}

heta_1 = \sin^{-1}\left(\frac{3}{4}ight)

Question 9

A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 	imes 10^{-5} cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is

Accepted Answer

0.15 cm. The first minimum is formed at a distance Y = \frac{\lambda D}{a}, where D is the focal length of the lens.

Y = \frac{(5 	imes 10^{-5}) 	imes 60}{0.02} = 0.15 cm

Question 10

For a parallel beam of monochromatic light of wavelength \lambda, diffraction is produced by a single slit whose width a is of the order of the wavelength of the light. If D is the distance of the screen from the slit, the width of the central maxima will be

Accepted Answer

\frac{2D\lambda}{a}. For the first minimum, \sin	heta = \frac{\lambda}{a}, and for small angles \sin	heta = \frac{Y}{D}

Y = \frac{\lambda D}{a}

Width of central maximum = 2Y = \frac{2\lambda D}{a}

Question 11

At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is

Accepted Answer

\pi radian. For the first minimum, a\sin	heta = \lambda.

The path difference between the wavelet from the edge and from the midpoint of the slit is \frac{a}{2}\sin	heta = \frac{\lambda}{2}

Phase difference = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{2} = \pi rad

Question 12

A beam of light of \lambda = 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central bright fringe is

Accepted Answer

2.4 mm. The distance between the first dark fringes on either side is the width of the central maximum:

x = \frac{2\lambda D}{a} = \frac{2 	imes 600 	imes 10^{-9} 	imes 2}{10^{-3}} = 2.4 	imes 10^{-3} m = 2.4 mm

Question 13

The angular resolution of a 10 cm diameter telescope at a wavelength of 5000 Å is of the order of

Accepted Answer

10^{-6} rad. Angular resolution of a telescope:

= \frac{1.22\lambda}{d} = \frac{1.22 	imes 5000 	imes 10^{-10}}{10 	imes 10^{-2}} = 6.1 	imes 10^{-6} \approx 10^{-6} rad

Diffraction and Polarisation of Light — NEET Physics PYQ MCQs with Solutions

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