In Young's double slit experiment, white light is used. The separ… - Vidyadip

MCQ

In Young's double slit experiment, white light is used. The separation between the slits is $ b. $ The screen is at a distance $d (d>> b)$ from the slits. Some wavelengths are missing exactly in front of one slit. These wavelengths are

A
$\lambda = \frac{{{b^2}}}{d}$
B
$\lambda = \frac{{2{b^2}}}{d}$
C
$\lambda = \frac{{{b^2}}}{{3d}}$
✓
Both $ (a) $ and $(c)$

✓

Answer

Correct option: D.

Both $ (a) $ and $(c)$

d
(d) Path difference between the rays reaching infront of slit $S_1$ is.
${S_1}P - {S_2}P = {({b^2} + {d^2})^{1/2}} - d$
For distructive interference at $P$
${S_1}P - {S_2}P = \frac{{(2\,n - 1)\lambda }}{2}$
i.e., ${({b^2} + {d^2})^{1/2}} - d = \frac{{(2n - 1)\lambda }}{2}$
$ \Rightarrow d\,{\left( {1 + \frac{{{b^2}}}{{{d^2}}}} \right)^{1/2}} - d = \frac{{(2n - 1)\lambda }}{2}$
$ \Rightarrow d\,\left( {1 + \frac{{{b^2}}}{{2{d^2}}} + ......} \right) - d = \frac{{(2n - 1)\lambda }}{2}$
(Binomial Expansion)
$ \Rightarrow \frac{b}{{2d}} = \frac{{(2n - 1)\lambda }}{2} \Rightarrow \lambda = \frac{{{b^2}}}{{(2n - 1)d}}$
For $n = 1,\;2............,\;\lambda = \frac{{{b^2}}}{d},\;\frac{{{b^2}}}{{3d}}$

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