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

The optical path difference needed for destructive interference is
$$2\mu d = \left( {2n + 1} \right)\frac{\lambda }{2},n = 0,1,2,......$$
Note that $$2\mu d$$  is the total optical path length that the rays traverse when $$n = 0.$$
$$\eqalign{ & \therefore d = \frac{{\frac{\lambda }{2}}}{{2\mu }} = \frac{\lambda }{{4\mu }} = \frac{{350 \times {{10}^{ - 9}}}}{{4 \times 1.38}} \cr & = 100\,nm = 1 \times {10^{ - 7}}m \cr} $$

For constructive interference $$d\sin \theta = n\lambda $$
given $$d = 2\,\lambda $$
$$ \Rightarrow \,\,\sin \theta = \frac{n}{2}$$
$$n = 0, 1, - 1, 2, - 2$$     hence five maxima are possible

Optical path difference
$$\Delta x = \left( {{\mu _2} - {\mu _1}} \right)t.$$

Treating the anti-reflectance coating like a camera-lens coating, one can obtain for its thickness $$t$$ :
$$2t = \left( {m + \frac{1}{2}} \right)\frac{\lambda }{n},m = 0, \pm 1, \pm 2,....\,\,......\left( {\text{i}} \right)$$
Condition (1) is the condition for the destructive interference for the normally incident and reflected electromagnetic wave. The minimum thickness of the coat refers to $$m = 0.$$  This gives for the thickness :
$$t = \frac{\lambda }{{4n}} = \frac{{3.00}}{{4\left( {1.5} \right)}}cm = 0.500\,cm.$$
This anti-reflectance coating could be easily countered by changing the wavelength of the radar to $$1.50\,cm$$  now creating maximum reflection.

The path difference produced by a sheet
$$\Delta = \left( {\mu - 1} \right)t$$
According to the given condition (for minimum thickness)
$$\eqalign{ & \left( {\mu - 1} \right)t = \frac{\lambda }{2} \cr & \therefore t = \frac{\lambda }{{2\left( {\mu - 1} \right)}} = \frac{{6600 \times {{10}^{ - 10}}}}{{2\left( {1.6 - 1} \right)}} = 5500\,\mathop {\text{A}}\limits^ \circ \cr} $$

We know that
$$\eqalign{ & I\left( \theta \right) = {I_0}{\cos ^2}\frac{\delta }{2}\,\,{\text{where}}\,\,\delta = \frac{{2\pi d\tan \theta }}{\lambda } \cr & I\left( \theta \right) = {I_0}{\cos ^2}\left( {\frac{{\pi d\tan \theta }}{\lambda }} \right) = {I_0}{\cos ^2}\left( {\frac{{\pi \times 150 \times \tan \theta }}{{\frac{{3 \times {{10}^8}}}{{{{10}^6}}}}}} \right) \cr & = {I_0}{\cos ^2}\left( {\frac{\pi }{2}\tan \theta } \right) \cr} $$
$$\eqalign{ & {\text{For}}\,\,\theta = {30^ \circ }; \cr & I\left( \theta \right) = {I_0}{\cos ^2}\left( {\frac{\pi }{{2\sqrt 3 }}} \right) \cr & {\text{For}}\,\,\theta = {90^ \circ };I\left( \theta \right) = {I_0}{\cos ^2}\left( \infty \right) \cr & {\text{For}}\,\,\theta = {0^ \circ };I\left( \theta \right) = {I_0} \cr} $$
$$I\left( \theta \right)$$  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.

Initially the parallel beam is cylindrical . Therefore, the wave front will be planar.

Axis of transmission of $$A$$ & $$B$$ are parallel.

After introducing polariser $$C$$ between $$A$$ and $$B,$$

$$\eqalign{ & \frac{I}{2}{\cos ^4}\theta = \frac{I}{8} \cr & \Rightarrow \,\,{\cos ^4}\theta = \frac{1}{4} \cr & \Rightarrow \,\,\cos \theta = \frac{1}{{\sqrt 2 }} \cr & {\text{or, }}\theta = {45^ \circ } \cr} $$

$$\eqalign{ & \frac{{12\,{\lambda _1}D}}{d} = \frac{{k{\lambda _2}D}}{d}, \cr & k = \frac{{12 \times 600}}{{400}} \cr & = 18 \cr} $$

$$\eqalign{ & n = \frac{{\left( {\mu - 1} \right)tD}}{d}\,\,{\text{but}}\,\,\beta = \frac{{\lambda D}}{d} \cr & \Rightarrow \frac{D}{d} = \frac{\beta }{\lambda } \cr & n = \frac{{\left( {\mu - 1} \right)t\beta }}{\lambda } \cr & 20\beta = \left( {\mu - 1} \right)2.5 \times {10^{ - 3}}\left( {\frac{\beta }{{5000 \times {{10}^{ - 8}}}}} \right) \cr & \mu - 1 = \frac{{20 \times 5000 \times {{10}^{ - 8}}}}{{2.5 \times {{10}^{ - 3}}}} \cr & \Rightarrow \mu = 1.4 \cr} $$