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Physics NEET Questions — Free MCQ Practice

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In a single slit diffraction of light of wavelength λ by a slit of width e, the size of the central maximum on a screen at a distance b is

A.

$2 b λ + e$

B.

$\frac{2 b λ}{e}$

C.

$\frac{2 b λ}{e} + e$

D.

$\frac{2 b λ}{e} - e$

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The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit is

A.

1 : 4 : 9

B.

1 : 2 : 3

C.

$1 : \frac{4}{9 π^{2}} : \frac{4}{25 π^{2}}$

D.

$1 : \frac{1}{π^{2}} : \frac{9}{π^{2}}$

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Explanation

$I = I_{0} {[\frac{\sin α}{α}]}^{2},$ where $α = \frac{ϕ}{2}$

For n^th secondary maxima $d \sin θ = (\frac{2 n + 1}{2}) λ$

$\Rightarrow α = \frac{ϕ}{2} = \frac{π}{λ} [d \sin θ] = (\frac{2 n + 1}{2}) π$

$∴ I = I_{0} {[\frac{\sin (\frac{2 n + 1}{2}) π}{(\frac{2 n + 1}{n}) π}]}^{2} = \frac{I_{0}}{{\{\frac{(2 n + 1)}{2} π\}}^{2}}$

So $I_{0} : I_{1} : I_{2} = I_{0} : \frac{4}{9 π^{2}} I_{0} : \frac{4}{25 π^{2}} I_{0}$

$= 1 : \frac{4}{9 π^{2}} : \frac{4}{25 π^{2}}$

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In a YDSE bi-chromatic light of wavelengths, 400 nm and 560 nm are used. The distance between the slits is 0.1 mm and the distance between the plane of the slits and the screen is 1m. The minimum distance between two successive regions of complete darkness is

A.

4 mm

B.

5.6 mm

C.

14 mm

D.

28 mm

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Explanation

Let nth minima of 400 nm coincides with mth minima of 560 nm then,

$(2 n - 1) 400 = (2 m - 1) 560$ ⇒ $\frac{2 n - 1}{2 m - 1} = \frac{7}{5} = \frac{14}{10} = \frac{21}{15}$

$F o r \frac{2 n - 1}{2 m - 1} = \frac{7}{5}; \Rightarrow 2 n - 1 = 7 \Rightarrow n = 4 \Rightarrow 2 m - 1 = 5 \Rightarrow m = 3$

i.e. 4th minima of 400 nm coincides with 3rd minima of 560 nm.

The location of this minima is

$= \frac{7 (1000) (400 \times 10^{- 6})}{2 \times 0 .1} = 14 m m$

Next, 11th minima of 400 nm will coincide with 8th minima of 560 nm

Location of this minima is;

$= \frac{21 (1000) (400 \times 10^{- 6})}{2 \times 0 .1} = 42 m m$

∴ Required distance = 28 mm

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In Young's double-slit experiment, the intensity at a point is (1/4) of the maximum intensity. Angular position of this point is :

A.

sin^-1(λ/d)

B.

sin^-1(λ/2d)

C.

sin^-1(λ/3d)

D.

sin^-1(λ/4d)

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Explanation

$I = \frac{I_{0}}{4} = I_{0} \cos^{2} (∆ ϕ / 2) \Rightarrow \cos^{2} (∆ ϕ / 2) = \frac{1}{4} \Rightarrow ∆ ϕ = 2 π / 3 So, Δx \times (2 π / λ) = ∆ ϕ = 2 π / 3 Δx = λ / 3 For angular position θ on screen, \sin θ = \frac{Δx}{d} \sin θ = \frac{λ}{3 d} \Rightarrow θ - si n^{- 1} (\frac{λ}{3 d})$

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A beam of electron is used in a YDSE experiment. The slit width is d. When the velocity of the electron is increased, then,

A.

No interference is observed

B.

Fringe width increases

C.

Fringe width decreases

D.

Fringe width remains the same

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Explanation

Momentum of the electron will increase. So the wavelength ( $λ$ = h/p) of electrons will decrease and fringe width decreases as β ∝ λ.

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If the separation between screen and source is increased by 2% what would be the effect on the intensity :

A.

Increases by 4%

B.

Increases by 2%

C.

Decreases by 2%

D.

Decreases by 4%

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Explanation

$I \propto \frac{1}{r^{2}}$ ⇒ $\frac{Δ I}{I} = - 2 \frac{Δ r}{r}$ = $- 2 \times 2 = - 4 %$

Hence intensity is decreased by 4%.

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The maximum kinetic energy of photoelectron emitted from the surface of work function f due to incidence of light of frequency n is E. If the frequency of incident light is doubled, then maximum kinetic of emitted photon will be

A.

2E

B.

2E - f

C.

2E + f

D.

2E + 2f

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Explanation

$KE = hν - ϕ E = hn - f When the frequency is doubled; (K . E .) = 2 hn - f = 2 hn - 2 f + f = 2 (hn - f) + f (K . E .)' = 2 E + f$

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The de-Broglie associated with an electron accelerated through a voltage of 900 V is:

A.

$0.31 \overset{o}{A}$

B.

$0.41 \overset{o}{A}$

C.

$0.5 \overset{o}{A}$

D.

$0.16 \overset{0}{A}$

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Explanation

$λ = \sqrt{\frac{150}{V}} = \sqrt{\frac{150}{900}} = \sqrt{\frac{1}{6}} = 0.41 \overset{o}{A}$

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$A 200 W sodium street lamp emits yellow light of wavelength 0.60 μm . Assuming it to be 50 % efficient, in converting electrical energy into light, the number of photons of yellow light emitted per second is nearly equal to -$

A.

$3 x 10^{19}$

B.

$3 x 10^{20}$

C.

$1.5 x 10^{20}$

D.

$6 x 10^{20}$

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Explanation

$(2) P = \frac{n}{t} \frac{h c}{λ} \frac{200 x 50}{100} = \frac{n}{1} \times \frac{6.6 x 10^{- 34} x 3 x 10^{8}}{6 x 10^{- 7}} n = 3 x 10^{20} p e r s e c$

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In a photoelectric experiment using a metal of work function 1.8 eV, if the maximum kinetic energy of emitted electrons is 1.5eV, then the corresponding value to stopping potential is:

A.

1.8V

B.

3.3V

C.

0.3V

D.

1.5V

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Explanation

$(4) Stopping potential = \frac{{(K . E .)}_{\max}}{e} = 1.5 V$

Physics MCQs for NEET — Practice Questions with Answers

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