Wave Optics MCQ Questions & Answers in Optics and Wave | Physics

A slit would give divergent; a biprism would give double; a glass slab would give a parallel wavefront. Edge is downward.

The shape of interference fringes formed on a screen in case of a monochromatic source is a straight line.
Remember for double hole experiment a hyperbola is generated.

$$\sin \theta = \frac{{0.25}}{{25}} = \frac{1}{{100}}$$

Resolving power $$ = \frac{{1.22\lambda }}{{2\mu \sin \theta }} = 30\,\mu m.$$

Shift $$ = \frac{{\beta \left( {\mu - 1} \right)}}{\lambda }{t_1} - \frac{{\beta \left( {\mu - 1} \right)}}{\lambda }{t_2}$$
$$ = \frac{{\beta \left( {\mu - 1} \right)}}{\lambda }\left( {{t_1} - {t_2}} \right)$$

We know that
$$\eqalign{ & I\left( \theta \right) = {I_0}{\cos ^2}\frac{\delta }{2}\,\,{\text{where }}\delta = \frac{{2\,\pi d\tan \theta }}{\lambda } \cr & I\left( \theta \right) = {I_0}{\cos ^2}\left( {\frac{{\pi d\tan \theta }}{\lambda }} \right) \cr & = {I_0}{\cos ^2}\left( {\frac{{\pi \times 150 \times \tan \theta }}{{3 \times \frac{{{{10}^8}}}{{{{10}^6}}}}}} \right) \cr & = {I_0}{\cos ^2}\left( {\frac{\pi }{2}\tan \theta } \right) \cr} $$

For $$\theta = {30^ \circ };I\left( \theta \right) = {I_0}{\cos ^2}\left( {\frac{\pi }{{2\sqrt 3 }}} \right)$$
For $$\theta = {90^ \circ };I\left( \theta \right) = {I_0}{\cos ^2}\left( \infty \right)$$
For $$\theta = {0^ \circ }$$
$$I\left( \theta \right) = {I_0}$$
$$I\left( \theta \right)$$  is not constant.
Alternatively, when $$\theta $$ is zero the path difference between wave originating from $${S_1}$$ and that from $${S_2}$$ will be zero. This corresponds to a maxima.

$$\eqalign{ & \beta = \frac{\lambda }{{2\mu \tan \alpha }} \simeq \frac{\lambda }{{2\mu \alpha }} \cr & \therefore \lambda = \beta \times 2\mu \alpha \cr & = 0.12 \times {10^{ - 2}} \times 2 \times 1 \times \left( {\frac{{40}}{{60 \times 60}} \times \frac{\pi }{{180}}} \right) \cr & = 4655\,\mathop {\text{A}}\limits^ \circ . \cr} $$

Use relativistic doppler's effect as velocity of observer is not small as compared to light
$$f = {f_0}\sqrt {\frac{{c + v}}{{c + v}}} ;$$
$$V$$ = relative speed of approach
$${f_0} = 10\,GHz$$
$$\eqalign{ & \therefore \,\,f = 10\sqrt {\frac{{c + \frac{c}{2}}}{{c - \frac{c}{2}}}} \cr & = 10\sqrt 3 \cr & = 17.3\,GHz \cr} $$

Path difference $$ = \left( {\mu - 1} \right)t = n\lambda ;$$
For minimum $$t,n = 1;$$
$$\therefore t = 2\,\lambda $$

$$\because $$ The E.M. wave are transverse in nature i.e.,
$$\eqalign{ & = \frac{{\overrightarrow k \times \overrightarrow E }}{{\mu \omega }} = \overrightarrow H \,\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{where, }}\overrightarrow H = \frac{{\overrightarrow B }}{\mu } \cr & {\text{and, }}\frac{{\overrightarrow k \times \overrightarrow H }}{{\omega \varepsilon }} = - \overrightarrow E \,\,\,\,.....\left( {{\text{ii}}} \right) \cr & \overrightarrow k \,\,{\text{is }} \bot \,\,\overrightarrow H \,\,{\text{and }}\overrightarrow k \,\,{\text{is also }} \bot \,\,{\text{to }}\overrightarrow E \cr} $$
or In other words $$\overrightarrow X \parallel \overrightarrow E \,\,{\text{and }}\overrightarrow k \parallel \overrightarrow E \times \overrightarrow B $$

$$\eqalign{ & {\text{Let, }}{I_1} = I\,\,{\text{and }}{I_2} = 4I \cr & {I_{\max }} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2} \cr & = {\left( {\sqrt I + \sqrt {4I} } \right)^2} \cr & = {\left( {3\sqrt I } \right)^2} \cr & = 9I \cr & {I_{\min }} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2} \cr & = {\left( {\sqrt I - \sqrt {4I} } \right)^2} \cr & = I \cr} $$