Optics 6 Question 54
56. Angular width of central maximum in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength $6000 \AA$. When the slit is illuminated by light of another wavelength, the angular width decreases by $30 %$. Calculate the wavelength of this light. The same decrease in the angular width of central maximum is obtained when the original apparatus is immersed in a liquid. Find refractive index of the liquid.
$(1996,2 M)$
Show Answer
Answer:
Correct Answer: 56. $4200 \AA, 1.43$
Solution:
- (a) Given, $\lambda=6000 \AA$
Let $b$ be the width of slit and $D$ the distance between screen and slit.
First minima is obtained at $b \sin \theta=\lambda$
or $b \theta=\lambda$ as $\sin \theta \approx \theta$ or $\theta=\frac{\lambda}{b}$
Angular width of first maxima $=2 \theta=\frac{2 \lambda}{b} \propto \lambda$
Angular width will decrease by $30 %$ when $\lambda$ is also decreased by $30 %$.
Therefore, new wavelength
$$ \begin{aligned} & \lambda^{\prime}=(6000)-\frac{30}{100} 6000 \AA \\ & \lambda^{\prime}=4200 \AA \end{aligned} $$
(b) When the apparatus is immersed in a liquid of refractive index $\mu$, the wavelength is decreased $\mu$ times. Therefore,
$$ \begin{aligned} & 4200 \AA & =\frac{6000 \AA}{\mu} \\ & \mu & =\frac{6000}{4200} \\ \text { or } & \mu & =1.429 \simeq 1.43 \end{aligned} $$
