Optics Ques 176
- A beam of light consisting of two wavelengths, $6500 $ $\AA$ and $5200 $ $\AA$ is used to obtain interference fringe in a Young’s double slit experiment.
$(1985,6$ M)
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength $6500$ $ \AA$.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
The distance between the slits is $2$ $ mm$ and the distance between the plane of the slits and the screen is $120 $ $cm$.
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Answer:
Correct Answer: 176.(a) $1.17$ $mm$ (b)$1.56$ $ mm$
Solution:
- (a) The desired distance will be
$ \begin{aligned} y _1 & =3 \omega _1=3 (\frac{\lambda _1 D}{d})=\frac{(3)\left(6500 \times 10^{-10}\right)(1.2)}{\left(2 \times 10^{-3}\right)} \\ & =11.7 \times 10^{-4} m=1.17 mm \end{aligned} $
(b) Let $n _1$ bright fringe of $\lambda _1$ coincides with $n _2$ bright fringe of $\lambda _2$. Then,
$\frac{n _1 \lambda _1 D}{d}=\frac{n _2 \lambda _2 D}{d}$ or $\frac{n _1}{n _2}=\frac{\lambda _2}{\lambda _1}=\frac{5200}{6500}=\frac{4}{5}$
Therefore, $4th$ bright of $\lambda _1$ coincides with $5th$ bright of $\lambda _2$. Similarly, $8th$ bright of $\lambda _1$ will coincide with $10th$ bright of $\lambda _2$ and so on.
The least distance from the central maximum will therefore corresponding to $4 th$ bright of $\lambda _1$ (or $5 th$ bright of $\lambda _2$.) Hence,
$ \begin{aligned} Y _{\text {min }} & =\frac{4 \lambda _1 D}{d}=\frac{4\left(6500 \times 10^{-10}\right)(1.2)}{\left(2 \times 10^{-3}\right)} \\ & =15.6 \times 10^{-4} m=1.56 mm \end{aligned} $
BETA
