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A slit of width a is illuminated by white light. For red light (λ = 6500 Å), the first minima is obtained at θ = 30°. Then the value of a will be

3250 Å

6.5 × 10^–4 mm

1.24 microns

2.6 × 10^–4 cm

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Explanation

For first minima $θ = \frac{λ}{a}$ or $a = \frac{λ}{θ}$

$∴ a = \frac{6500 \times 10^{- 8} \times 6}{π}$ (As 30^o = $\frac{π}{6}$ radian)

$= 1 .24 \times 10^{- 4} c m = 1 .24$ microns

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The radius of central zone of the circular zone plate is 2.3 mm. The wavelength of incident light is $5893 Å .$ Source is at a distance of 6m. Then the distance of the first image will be

9 m

12 m

24 m

36 m

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Explanation

$f_{1} = \frac{r^{2}}{λ} = \frac{{(2 .3 \times 10^{- 3})}^{2}}{5893 \times 10^{- 10}} = 9 m .$

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What will be the angular width of central maxima in Fraunhoffer diffraction when light of wavelength $6000 Å$ is used and slit width is 12×10^–5 cm

2 rad

3 rad

1 rad

8 rad

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Explanation

Angular width $= \frac{2 λ}{d} = \frac{2 \times 6000 \times 10^{- 10}}{12 \times 10^{- 5} \times 10^{- 2}} = 1 r a d .$

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Direction of the first secondary maximum in the Fraunhofer diffraction pattern at a single slit is given by (a is the width of the slit)

$a \sin θ = \frac{λ}{2}$

$a \cos θ = \frac{3 λ}{2}$

$a \sin θ = λ$

$a \sin θ = \frac{3 λ}{2}$

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Explanation

For n^th secondary maxima path difference

$d \sin θ = (2 n + 1) \frac{λ}{2} \Rightarrow a \sin θ = \frac{3 λ}{2}$

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A parallel beam of monochromatic light of wavelength 5000 Å is incident normally on a single narrow slit of width 0.001 mm. The light is focused by a convex lens on a screen placed on the focal plane. The first minimum will be formed for the angle of diffraction equal to

0^o

15^o

30^o

60^o

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Explanation

For the first minima $d \sin θ = λ$

$\Rightarrow \sin θ = \frac{λ}{d} \Rightarrow θ = \sin^{- 1} (\frac{5000 \times 10^{- 10}}{0 .001 \times 10^{- 3}}) = 30^{o}$

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In the far field diffraction pattern of a single slit under polychromatic illumination, the first minimum with the wavelength $λ_{1}$ is found to be coincident with the third maximum at $λ_{2}$ . So

$3 λ_{1} = 0 .3 λ_{2}$

$3 λ_{1} = λ_{2}$

$λ_{1} = 3 .5 λ_{2}$

$0 .3 λ_{1} = 3 λ_{2}$

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Explanation

Position of first minima = position of third maxima i.e., $\frac{1 \times λ_{1} D}{d} = \frac{(2 \times 3 + 1)}{2} \frac{λ_{2} D}{d} \Rightarrow λ_{1} = 3 .5 λ_{2}$

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The angle of polarisation for any medium is 60^o, what will be critical angle for this

$\sin^{- 1} \sqrt{3}$

$\tan^{- 1} \sqrt{3}$

$\cos^{- 1} \sqrt{3}$

$\sin^{- 1} \frac{1}{\sqrt{3}}$

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Explanation

By using $μ = \tan θ_{p} \Rightarrow μ = \tan 60 = \sqrt{3}$ ,

also $C = \sin^{- 1} (\frac{1}{μ}) \Rightarrow C = \sin^{- 1} (\frac{1}{\sqrt{3}})$

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In the propagation of electromagnetic waves, the angle between the direction of propagation and plane of polarisation is

0^o

45^o

90^o

180^o

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Explanation

Plane of polarization is a confinement of the electric/magnetic field vector to a given plane along the direction of propagation. Therefore the angle between them is $0^{0}$

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A light has amplitude A and the angle between analyzer and polariser is 60°. Light reflected by analyzer has amplitude

$A \sqrt{2}$

$A / \sqrt{2}$

$\sqrt{3} A / 2$

A/2

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Explanation

The amplitude will be $A \cos 60^{o} = A / 2$

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Light passes successively through two polarimeters tubes each of length 0.29m. The first tube contains dextro rotatory solution of concentration 60kgm^–3 and specific rotation 0.01rad m²kg^–1. The second tube contains laevo rotatory solution of concentration 30kg/m³ and specific rotation 0.02 radm²kg^–1. The net rotation produced is

15°

0°

20°

10°

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Explanation

Rotation produced θ = Slc

Net rotation produced θ_r = θ₁ – θ₂ = l (S₁c₁ – S₂c₂)

= 0.29 × [0.01 × 60 – 0.02 × 30] = $0^{0}$

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