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NEET NEET Questions — Free MCQ Practice

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Light of waveâ€“length $ \lambda $ is incident on a slit of width d. The resulting diffraction pattern is observed on a screen placed at a distance D. The linear width of the principal maximum is equal to the width of the slit, then D = ______.

A. $ { d^2 \over 2 \lambda } $ B. $ { 2 \lambda^2 \over d } $ C. $ { d \over \lambda } $ D. $ { 2 \lambda \over d } $

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Explanation

The linear width of the principal maximum in a single-slit diffraction pattern is given by $2\lambda D / d$. If this width is equal to the width of the slit (d), then we can set up the equation $2\lambda D / d = d$. Solving for D, we get $D = \frac{d^2}{2\lambda}$. Hence, the correct option is $\frac{d^2}{2\lambda}$.

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A polariser is used for

A. Produced polarised light B. Produced unpolarised light C. Produced unpolarised light D. None of these

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Explanation

A polarizer is a device that converts unpolarized light into polarized light by allowing only light waves oscillating in a particular direction to pass through. Hence, the correct option is 'Produced polarised light'.

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Read the paragraph and chose the correct answer of the following questions In young experiment position of bright fringes is given $ x = n \lambda { D \over d } $ and the positon of dark fringes is given by $ x = (2n -1 ) { \lambda \over 2 } { D \over d } $ where n = 1,2,3........... for first, second, third bright / dark fringe. The center of the fringe pattern is bright (for n = 0). The width of each briht/dark fringe is $ \lambda = 5000 A ^\circ $ . If light of wavelength $6000 A^\circ $ be used in the above experiment the fringe width would be ........mm

A. 0.36 B. 3 C. 0.6 D. 6

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Read the paragraph and chose the correct answer of the following questions In young experiment position of bright fringes is given $ x = n \lambda { D \over d } $ and the positon of dark fringes is given by $ x = (2n -1 ) { \lambda \over 2 } { D \over d } $ where n = 1,2,3........... for first, second, third bright / dark fringe. The center of the fringe pattern is bright (for n = 0). The width of each briht/dark fringe is $ \lambda = 5000 A ^\circ $ . With the light of wavelength $ 5000 A^\circ $ , If experiment were carried out under water of a n = 4 /3 the fringe width would be

A. zero B. 4/3 times C. 3/4 times D. none of these

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Explanation

When the experiment is carried out in a medium with refractive index $ n $, the wavelength of light changes to $ \lambda' = \frac{\lambda}{n} $. Therefore, the new fringe width $ \beta' $ is given by $ \beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{nd} $. Thus, the fringe width in water (where $ n = \frac{4}{3} $) will be $ \frac{3}{4} $ times the fringe width in air.

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The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing through its centre and perpendicular to its plane will be :

A.

$ I_1 $ + $I_2$

B.

$ I_1 $ - $I_3$

C.

$ I_1 $ + $I_2$ + $I_4$

D.

$ I_1 $ + $I_2$ + $I_3$

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Explanation

Use perpendicular axis theorem

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In a fraunhofer diffraction by single slit of width d with incident light of wavelength $ 5500 A ^\circ $ the first minimum is observed at angle of $30 ^\circ $ . The first secondary maximum is observed at an angle $ \theta $ =

A. $ sin ^{-1} \left( { 1 \over \sqrt 2} \right) $ B. $ sin ^{-1} \left( { 3 \over 4 } \right) $ C. $ sin ^{-1} \left( { \sqrt 3 \over 2 } \right) $ D. $ sin ^{-1} \left( { 1 \over 4 } \right) $

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The phenomenon of polarisation of electromagnetic waves proves that the electromagnetic waves are

A. mechanical B. longitudial C. transverse D. none of these

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Explanation

The phenomenon of polarization is characteristic of transverse waves, not longitudinal waves. In polarization, the oscillations of the electromagnetic wave are restricted to a particular direction perpendicular to the direction of wave propagation. This can only happen if the waves are transverse. Therefore, polarization proves that electromagnetic waves are transverse in nature.

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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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When the angle of incidence is $ 60 ^\circ $ on the Surface of a glass slab, it is found that the reflected ray is completely palarised. Then the velocity of light in glass is____$ ms^{-1} $

A. $ \sqrt 2 $ B. $ \sqrt 3 \times 10^8 $ C. $ \sqrt 3 $ D. $ \sqrt 2 \times 10^8 $

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Explanation

$ Here , n = tan \theta_p = \sqrt 3 , n = {c \over \nu } $ $ \therefore v = \sqrt 3 \times 10^8 $

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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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