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

Angular width of central maxima $$ = \frac{{2\lambda }}{d}$$
$$\eqalign{ & {\text{or, }}\lambda = \frac{d}{2};{\text{Fringe width,}}\beta = \frac{{\lambda \times D}}{{d'}} \cr & {10^{ - 2}} = \frac{d}{2} \times \frac{{50 \times {{10}^{ - 2}}}}{{d'}} \cr & = \frac{{{{10}^{ - 6}} \times 50 \times {{10}^{ - 2}}}}{{2 \times d'}} \cr} $$
Therefore, slit separation distance, $$d' = 25\,\mu m$$

$$\eqalign{ & \mu = \tan i \cr & \Rightarrow i = {\tan ^{ - 1}}\left( \mu \right) = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) = {60^ \circ }. \cr} $$

Locus of equal path difference are lines running parallel to axis of the cylinder. Hence straight fringes will be observed.

The intensity $$I$$ is given as
$$I = {I_0}{\cos ^2}\frac{f}{2}$$
where $${I_0}$$ is the peak intensity
Here, $$I = \frac{{{I_0}}}{2},$$
$$\eqalign{ & \therefore \,\,\frac{{{I_0}}}{2} = {I_0}{\cos ^2}\frac{\phi }{2} \cr & \therefore \,\,\phi = \frac{\pi }{2}\left( {2\,n + 1} \right) \cr} $$
For a phase difference of $$2\,\pi $$  the path difference is $$\lambda $$
∴ For a phase difference of $$\left( {2\,n + 1} \right)\frac{p}{2}$$   the path difference is
$$\left( {2\,n + 1} \right)\frac{\lambda }{4}.$$
option (B) is correct.

$$\eqalign{ & {y_n} = \frac{{nD{\lambda _1}}}{d} = \frac{{\left( {n + 1} \right)D{\lambda _2}}}{d} \cr & {\text{or}}\,\,n = \left( {\frac{{{\lambda _2}}}{{{\lambda _2} - {\lambda _1}}}} \right) \cr & {\text{and}}\,\,{y_n} = \left( {\frac{{{\lambda _1}{\lambda _2}}}{{{\lambda _2} - {\lambda _1}}}} \right)\left( {\frac{D}{d}} \right). \cr} $$

NOTE : Frequency does not change with change of medium.

From $$\mu = \frac{c}{v} = \frac{{n{\lambda _v}}}{{n{\lambda _m}}},{\lambda _m} = \frac{{{\lambda _v}}}{\mu }$$
Here, $$c =$$  velocity of light in vacuum and
$$v =$$  velocity of light in medium
$$m =$$  refractive index of the medium.
Hence, wavelength in medium $$\left( {{\lambda _m}} \right) < {\lambda _v}\left( {\because \mu > 1,{\text{given}}} \right)$$
So, the required wavelength decreases.

$$\eqalign{ & \mu = \tan {i_p} = \tan {60^ \circ } = \sqrt 3 \cr & \sin {i_C} = \frac{1}{\mu } = \frac{1}{{\sqrt 3 }} \Rightarrow {i_C} = {\sin ^{ - 1}}\frac{1}{{\sqrt 3 }} \cr} $$

At the area of total darkness minima will occur for both the wavelengths.
$$\eqalign{ & \therefore \,\,\frac{{\left( {2\,n + 1} \right)}}{2}{\lambda _1} = \frac{{\left( {2\,m + 1} \right)}}{2}{\lambda _2} \cr & \Rightarrow \,\,\left( {\,2n + 1} \right){\lambda _1} = \left( {2\,m + 1} \right){\lambda _2} \cr & {\text{or, }}\frac{{\left( {2\,n + 1} \right)}}{{\left( {2\,m + 1} \right)}} = \frac{{560}}{{400}} = \frac{7}{5} \cr & {\text{or, }}10\,n = 14\,m + 2 \cr} $$
by inspection for $$m = 2, n = 3$$   and for $$m = 7, n = 10,$$   the distance between them will be the distance between such points.
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.,}}\,\,\,\Delta s = \frac{{D\,{\lambda _1}}}{d}\left\{ {\frac{{\left( {2\,{n_2} + 1} \right) - \left( {2\,{n_1} + 1} \right)}}{2}} \right\} \cr & {\text{put, }}{n_2} = 10,\,{n_1} = 3 \cr} $$
On solving we get,
$$\Delta s = 28\,mm.$$

In absence of film or for $$m = 0$$  intensity is maximum at screen. As the value of $$m$$ is increased, intensity shall decrease and then increase alternately. Hence the correct variation is (C).