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.

$(1996,2 M)$

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Answer:

Correct Answer: 56. $4200 \AA, 1.43$

Solution:

  1. (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,

$$ \begin{aligned} & 4200 \AA & =\frac{6000 \AA}{\mu} \\ & \mu & =\frac{6000}{4200} \\ \text { or } & \mu & =1.429 \simeq 1.43 \end{aligned} $$



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