Wave Optics - Result Question 22
22. In Young’s double slit experiment, the fringe width is found to be $0.4 mm$. If the whole apparatus is immersed in water of refrative index $\frac{4}{3}$, without disturbing the geometrical arrangement, the new fringe width will be [1990]
(a) $0.30 mm$
(b) $0.40 mm$
(c) $0.53 mm$
(d) 450 microns
Topic 3: Diffraction, Polarization of Light &
Resolving Power23. The Brewsters angle $i_b$ for an interface should be
[2020]
(a) $30^{\circ}<i_b<45^{\circ}$
(c) $i_b=90^{\circ}$
(b) $45^{\circ}<i_b<90^{\circ}$
(d) $0^{\circ}<i_b<30^{\circ}$
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Answer:
Correct Answer: 22. (a)
Solution:
(a) $\beta^{\prime}=\frac{\beta}{\mu}=\frac{0.4}{\frac{4}{3}}=0.3 mm$
If the whole YDSE set up is taken in another medium then $\lambda$ changes. So $B$ changes.
In water $\lambda \omega=\frac{\lambda_a}{\mu_w}$
$ \Rightarrow \quad \beta \omega=\frac{B_a}{\mu_w}=\frac{3}{4} B_a $
