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Wave Optics MCQs for NEET — Physics Questions with Answers

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Light from two coherent Sources of the same amplitude A and wavelength $ \lambda $ , illuminates the Screen. The intensity of the central maximum is Io. If the sources were incoherent, the intensity at the same point will be ____ __.

A. $ I_o / 2 $ B. $ I_o / 4 $ C. $ 4 I_o $ D. $2 I_o $

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Explanation

When two coherent sources interfere, the resulting intensity at any point on the screen is given by the principle of superposition. For coherent sources, the central maximum intensity is $I_o = 4A^2$. If the sources were incoherent, their intensities would simply add up because there is no fixed phase relationship between them. The intensity at the central point would then be the sum of the individual intensities of the two sources, which is $I_o / 2$.

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Two beams of Light of intensity $I_1 $ and $ I_2 $ . interfere to give an interference pattern. If the ratio of maximum intensity to that of minimum intensity is 16 /4 then $ { I_1 \over I_2 } $ = ..........

A. 1:9 B. 1:4 C. 4:1 D. 9:1

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Explanation

$from { I_{max} \over I_{max} } = { ( a+b)^2 \over ( a-b)^2 } $ $ 3b = 9 $ $ Now \therefore { I_1 \over I_2 } = { a^2 \over b^2 } = 9:1 $

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Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the resultant intensities at A and B is

A. I B. 4I C. 2I D. 6I

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Explanation

$ Here , I_A = I_1 + I_2 + 2 \sqrt { I_1 I_2 } $ $ cos { \pi \over 2 } = I \times 4 I \times 2 \sqrt { I \times 4 I } \times cos 90 ^\circ $ $ I_A = 5I $ $ and I_B = 5I + 2 \sqrt { I \times 4 I } \times cos \pi = 5 I -4 I = I $ $ \therefore I_A - I_B = 4 I $

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A sound source emits sound of 600 Hz frequency, this sound enters by opened door of width 0.75 m. Find the angle on one side at which fitst minimum is formed. The speed of sound = $300 ms ^{-1} $ .

A. $ 84.4 ^\circ $ B. $ 90 ^\circ $ C. $ 74.2 ^\circ $ D. $ 47.2 ^\circ $

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A plane polarised light is incident normally on the tourmaline plate. its $ \vec E $ vectors make an agnle of $ 45 ^\circ $ with the optical axis of the plate. find the percentage difference between intial and final maximum values of $ \vec E $ vectors.

A. $ 19 \% $ B. $ 92 \% $ C. $ 50 \% $ D. $ 29 \% $

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Explanation

$ I = I_0 cos ^2 \theta = { I_0 \over 2 } and { E^2 \over E_0^2 } = {1 \over 2} , { E \over E_0} = { 1 \over \sqrt 2 } $ $ \therefore { |E- E_0| \over E_0} = 0.29 = 29 \%$

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The ratio of intensities of rays emitted from two different coherent Sources is $ \lambda $ . . For the interference pattern by them , $ { Imax + Imin \over Imax -Imin } $ will be equal to ................

A. $ { 1 + \sqrt a \over 2 \alpha } $ B. $ { 1 + a \over 2 \alpha } $ C. $ { 1 + \sqrt a \over 2 } $ D. $ { 1 + \sqrt a \over 2 \sqrt \alpha } $

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Explanation

$ here { I_1 \over I_2 } = \alpha , \therefore {E_1 \over E_2 } = \sqrt \alpha $ $ and { E_1 + E_2 \over E_1 - E_2 } = { \sqrt \alpha + 1 \over \sqrt \alpha - 1 } $ $ \therefore { I_{max} \over I_{min} } = { (\sqrt \alpha + 1)^2 \over ( \sqrt \alpha -1 ) ^2 } $ $ \therefore { I _{max} + I_{min} \over I_{max} - I_{min} } = { 2 ( \alpha + 1 ) \over 4 \sqrt \alpha }= { \alpha +1 \over 2 \sqrt \alpha } $

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Assume that light of wavelength 7000 is coming from a star. The limit of resolution (in radian) of a telescope whose objective has a diameter of 244 cm, will be

$2.5 x 10^{- 7}$

$5.0 x 10^{- 7}$

$3.5 x 10^{- 7}$

$4.5 x 10^{- 7}$

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Explanation

$R . L . = \frac{1.22 λ}{a} = \frac{1.22 x 7 x 10^{- 7}}{2.44} R . L . = 3.5 x 10^{- 7}$

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We wish to see inside an atom. Assuming the atom to have a diameter of 100 pm, this means that one must be able to resolved a width of say 10 p.m. If an electron microscope is used, the minimum electron energy required is about

1.5 KeV

15 KeV

150 KeV

1.5 KeV

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Explanation

(b) Wave length of the electron wave be $10 \times 10^{- 12} m$ ,
Using $λ = \frac{h}{\sqrt{2 m E}} \Rightarrow E = \frac{h^{2}}{λ^{2} \times 2 m}$
$= \frac{{(6.63 \times 10^{- 34})}^{2}}{{(10 \times 10^{- 12})}^{2} \times 2 \times 9.1 \times 10^{- 31}} J o u l e$

$= \frac{{(6.63 \times 10^{- 34})}^{2}}{{(10 \times 10^{- 12})}^{2} \times 2 \times 9.1 \times 10^{- 31} \times 1.6 \times 10^{- 19}} e V = 15.1 K e V$

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Two point white dots are 1mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm. Approximately, what is the maximum distance at which dots can be resolved by the eye ? [Take wavelength of light = 500 nm]

6 m

3 m

5 m

1 m

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Explanation

(c)

$\frac{1.22 λ}{a} = \frac{x}{d} \Rightarrow d = \frac{x \times a}{1.22 λ} = \frac{1 \times 10^{- 3} \times 3 \times 10^{- 3}}{1.22 \times 500 \times 10^{- 9}} = 5 m$

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A parallel beam of moving electrons is incident normal on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the slit is further narrowed, then which of the following statement is correct?

The diffraction pattern is not observed on the screen in the case of electrons.

The angular width of the central maximum of the diffraction pattern will increase.

The angular width of the central maximum will decrease.

The angular width of the central maximum will remain the same.

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Explanation

$(2) β = \frac{2 λD}{a} If a ↓ then β ↑ .$

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