A thin paper of thickness $0.02 m m$ having refractive index $1.45$ is pasted across one of the slit in a Young’s double slit experiment. The paper transmits 4/9 of radient energy falling on it. (Wavelength $λ = 600 n m$ ).

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Amplitude of light wave transmitted through the paper will be 2/3 time of incident wave.

The ratio of maximum and minimum intensity in the pattern will be 5.

The ratio of maximum and minimum intensity in the fringe pattern will be 25.

The total number of fringes crossing the centre if an identical paper is pasted on the other slit is 15.

answer is A, B, C.

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Detailed Solution

$\frac{I_{\max}}{I_{\min}} = \frac{{(a_{1} + a_{2})}^{2}}{{(a_{1} - a_{2})}^{2}} a_{1} = \frac{2}{3} \sqrt{1} = \frac{2}{3} a_{0}$
$a_{2} = \sqrt{I} = a_{0}$
$∴ \frac{I_{\max}}{I_{\min}} = 25$
If an identical paper is pasted across second slit shifted C.B.F. will back in central point.
$s h i f t = (μ - 1) \frac{t D}{d}$
fringe width $= \frac{n λ D}{d}$
$∴ n = \frac{s h i f t}{f r i n g e w i d t h} = \frac{(μ - 1) t}{λ} = 15$