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  • In Young's double slit experiment, white light is used. The separation between the slits is b. The screen is at a distance <img alt="\mathrm{\mathrm{d}(\mathrm{d}>> b) }" src="https://ent

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

Option: 1

\mathrm{\lambda=\frac{\mathrm{b}^2}{2 \mathrm{~d}}}


Option: 2

\mathrm{\lambda=\frac{2 b^2}{d}}


Option: 3

\mathrm{\lambda=\frac{\mathrm{b}^2}{3 \mathrm{~d}}}


Option: 4

\mathrm{\lambda=\frac{2 b^2}{3 d}}


Answers (1)

best_answer

Path difference between the rays reaching in front of slit \mathrm{S}_1 is

\mathrm{S}_1 \mathrm{P}-\mathrm{S}_2 \mathrm{P}=\sqrt{\left(\mathrm{b}^2-\mathrm{d}^2\right)}-\mathrm{d}

For destructive interference at \mathrm{P}

\mathrm{ \mathrm{S}_1 \mathrm{P}-\mathrm{S}_2 \mathrm{P}=\frac{(2 \mathrm{n}-1) \lambda}{2}}
i.e., \mathrm{\quad \sqrt{\left(b^2+d^2\right)}-d=\frac{(2 n-1) \lambda}{2}}
\mathrm{ \begin{aligned} & \Rightarrow d \sqrt{\left(1+\frac{b^2}{d^2}\right)}-d=\frac{(2 n-1) \lambda}{2} \\ & \Rightarrow d\left(1+\frac{b^2}{2 d^2}\right)-d=\frac{(2 n-1) \lambda}{2} \\ & \Rightarrow \frac{\beta^2}{2 \delta}=\frac{(2 v-1) \lambda}{2} \\ & \therefore \lambda=\frac{b^2}{(2 n-1) d} \end{aligned} }

For \mathrm{\mathrm{n}=1,2, \ldots,}

\mathrm{ \Rightarrow \lambda=\frac{\mathrm{b}^2}{\mathrm{~d}}, \frac{\mathrm{b}^2}{3 \mathrm{~d}} \ldots \ldots }

Posted by

SANGALDEEP SINGH

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