IIT-JEE 2008 · Paper 1 · Physics Q33

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PhysicsMultiple correct+4 / -1Question 33 of 69

In a Young's double slit experiment, the separation between the two slits is $d$ and the wavelength of the light is $\lambda$ . The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s).

d=\lambda

, the screen will contain only one maximum

\lambda<d<2\lambda

, at least one more maximum (besides the central maximum) will be observed on the screen

If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase

If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase

Answer and worked solution

Official answer: A, B

Maxima. $d\sin\theta=m\lambda$ with $|\sin\theta|<1$ on the screen.

(A) $d=\lambda$ : $|m|<1$ , so only $m=0$ , the central maximum. True.
(B) $\lambda<d<2\lambda$ : $m=\pm1$ are also possible. True.

Fringe intensities with $I_1=4I$ , $I_2=I$ : bright $(\sqrt{4I}+\sqrt I)^2=9I$ , dark $(\sqrt{4I}-\sqrt I)^2=I$ .

(C) $I_1$ reduced to $I$ : bright $4I$ , dark 0. Both decrease. False.
(D) $I_2$ raised to $4I$ : bright $16I$ increases, but dark becomes 0, a decrease. False.

(A), (B)

Topics:Interference and Young's double slit

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IIT-JEE 2008 · Paper 1, Physics question 33