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Physics MCQs for NEET — Practice Questions with Answers

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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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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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When the angle of incidence on a material is 60°, the reflected light is completely polarized. The velocity of the refracted ray inside the material is (in ms^–1)

3 × 10⁸

$(\frac{3}{\sqrt{2}}) \times 10^{8}$

$\sqrt{3} \times 10^{8}$

0.5 × 10⁸

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Explanation

From Brewster's law

$μ = \tan i_{p}$ ⇒ $\frac{c}{v} = \tan 60 ° = \sqrt{3}$

⇒ $v = \frac{c}{\sqrt{3}} = \frac{3 \times 10^{8}}{\sqrt{3}} = \sqrt{3} \times 10^{8}$ m/sec.

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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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A circular disc is placed in front of a narrow source. When the point of observation is 2 m from the disc, then it covers first HPZ. The intensity at this point is I. When the point of observation is 25 cm from the disc then intensity will be

${(\frac{R_{6}}{R_{2}})}^{2} I$

${(\frac{R_{7}}{R_{2}})}^{2} I$

${(\frac{R_{8}}{R_{2}})}^{2} I$

${(\frac{R_{9}}{R_{2}})}^{2} I$

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Explanation

$I = \frac{R_{2}^{2}}{4}$ given $n_{1} b_{1} = n_{2} b_{2}$ ⇒ $1 \times 200 = n_{2} \times 25$

$∴ n_{2} = 8 H P Z$

$∴ {(\frac{R_{9}}{R_{2}})}^{2} I$

$= {(\frac{R_{9}}{R_{8}} \times \frac{R_{8}}{R_{7}} \times \frac{R_{7}}{R_{6}} \times \frac{R_{6}}{R_{5}} \times \frac{R_{5}}{R_{4}} \times \frac{R_{4}}{R_{3}} \times \frac{R_{3}}{R_{2}} \times \frac{R_{2}}{R_{2}})}^{2}$

$= {(\frac{R_{9}}{R_{2}})}^{2} I$

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