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Wave Optics MCQs for NEET — Physics Questions with Answers

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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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Unpolarized light falls on two polarizing sheets placed one on top of the other. What must be the angle between the characteristic directions of the sheets if the intensity of the final transmitted light is one-third the maximum intensity of the first transmitted beam?

$G i v e n : \cos 75^{0} = 0.26 \cos 57^{0} = 0.57 \cos 35^{0} = 0.82 \cos 15^{0} = 0.96$

75°

55°

35°

15°

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Explanation

$I' = \frac{I}{2} \cos^{2} θ = \frac{I}{6}$ or $\cos θ = \frac{1}{\sqrt{3}}$

∴ θ = 55º

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Unpolarized light of intensity 32Wm^–2 passes through three polarizers such that transmission axes of the first and second polarizer makes and angle 30° with each other and the transmission axis of the last polarizer is crossed with that of the first. The intensity of final emerging light will be

32 Wm^–2

3 Wm^–2

8 Wm^–2

4 Wm^–2

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Two polaroids are placed in the path of unpolarized beam of intensity I₀ such that no light is emitted from the second polaroid. If a third polaroid whose polarization axis makes an angle θ with the polarization axis of first polaroid, is placed between these polaroids then the intensity of light emerging from the last polaroid will be

$(\frac{I_{0}}{8}) \sin^{2} 2 θ$

$(\frac{I_{0}}{4}) \sin^{2} 2 θ$

$(\frac{I_{0}}{2}) \cos^{4} θ$

$I_{0} \cos^{4} θ$

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Explanation

The first two polaroids are arranged to block all light. When the third polaroid is introduced at an angle Î¸, the intensity transmitted is (Iâ‚€/8)sinÂ²(2Î¸). This is because the light has to go through three polarizers: the first transmits Iâ‚€/2, the second transmits (Iâ‚€/2)cosÂ²(Î¸), and the third transmits (Iâ‚€/2)cosÂ²(Î¸)sinÂ²(Î¸) = (Iâ‚€/8)sinÂ²(2Î¸).

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In the Young's double slit experiment, if the phase difference between the two waves interfering at a point is Ï•, the intensity at that point can be expressed by the expression-

(where A and B depend upon the amplitudes of the two waves)

$I = \sqrt{A^{2} + B^{2} \cos^{2} ϕ}$

$I = \frac{A}{B} \cos ϕ$

$I = A + B \cos \frac{ϕ}{2}$

$I = A + B \cos ϕ$

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Explanation

$I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ$

Put $I_{1} + I_{2} = A$ and $\sqrt{I_{1} I_{2}} = B; ∴ I = A + B \cos ϕ$

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When one of the slits of Young’s experiment is covered with a transparent sheet of thickness 4.8 mm, the central fringe shifts to a position originally occupied by the 30^th bright fringe. What should be the thickness of the sheet if the central fringe has to shift to the position occupied by 20^th bright fringe

3.8 mm

1.6 mm

7.6 mm

3.2 mm

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Explanation

If shift is equivalent to n fringes then

$n = \frac{(μ - 1) t}{λ} \Rightarrow n \propto t \Rightarrow \frac{t_{2}}{t_{1}} = \frac{n_{2}}{n_{1}} \Rightarrow t_{2} = \frac{n_{2}}{n_{1}} \times t$

$t_{2} = \frac{20}{30} \times 4 .8 = 3 .2 m m .$

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In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is

2λ

$\frac{2 λ}{3}$

$\frac{λ}{3}$

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Explanation

According to given condition

$(μ - 1) t = n λ$ for minimum t, n =1

So, $(μ - 1) t_{\min} = λ$

$t_{\min} = \frac{λ}{μ - 1} = \frac{λ}{1 .5 - 1} = 2 λ$

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In Young’s double slit experiment, the two slits act as coherent sources of equal amplitude A and wavelength λ. In another experiment with the same set up the two slits are of equal amplitude A and wavelength λ but are incoherent. The ratio of the intensity of light at the mid-point of the screen in the first case to that in the second case is

1 : 2

2 : 1

4 : 1

1 : 1

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Explanation

Resultant intensity $I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ$

At central position with coherent source (and $I_{1} = I_{2} = I_{0}$ )

$I_{c o n} = 4 I_{0}$ ... (i)

In case of incoherent at a given point, Ï• varies randomly with time so (cos Ï•)_av = 0

∴ $I_{I n c o h} = I_{1} + I_{2} = 2 I_{0}$ ... (ii)

Hence $\frac{I_{c o h}}{I_{I n c o h}} = \frac{2}{1}$ .

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