31.
An initially parallel cylindrical beam travels in a medium of refractive index $$\mu \left( I \right) = {\mu _0} + {\mu _2}I,$$ where $${\mu _0}$$ and $${\mu _2}$$ are positive constants and $$I$$ is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.
The initial shape of the wave front of the beam is
A
convex
B
concave
C
convex near the axis and concave near the periphery
Initially the parallel beam is cylindrical . Therefore, the wave front will be planar.
32.
A beam of light of wave length $$600\,nm$$ from a distance source falls on a single slit $$1mm$$ wide and a resulting diffraction pattern is observed on a screen $$2\,m$$ away. The distance between the first dark fringes on either side of central bright fringe is
The distance between the first dark fringe on either
side of the central maximum = width of central maximum
$$\eqalign{
& = \frac{{2\,D\lambda }}{a} \cr
& = \frac{{2 \times 2 \times 600 \times {{10}^{ - 9}}}}{{{{10}^{ - 3}}}} \cr
& = 2.4 \times {10^{ - 3}}m \cr
& = 2.4\,mm \cr} $$
33.
The correct curve between refractive index $$\mu $$ and wavelength $$\lambda $$ will be
According to Cauchy’s formula $$\mu = A + \frac{B}{{{\lambda ^2}}} + ......$$ Therefore, as $$\lambda $$ increases, $$\mu $$ decreases. Curve $$A$$ is correct.
34.
In an experiment of single slit diffraction pattern, first minimum for red light coincides with first maximum of some other wavelength, If wavelength of red light is $$6600\,\mathop {\text{A}}\limits^ \circ ,$$ then wavelength of first maximum will be :
In a single slit experiment,
For diffraction maxima, $$a\sin \theta = \left( {2n + 1} \right)\frac{\lambda }{2}$$
and for diffraction minima, $$a\sin \theta = n\lambda $$
According to question,
$$\eqalign{
& \left( {2 \times 1 + 1} \right)\frac{\lambda }{2} = 1 \times 6600\,\,\left( {\because {\lambda _R} = 6600\,\mathop {\text{A}}\limits^ \circ } \right) \cr
& \lambda = \frac{{6600 \times 2}}{3} = 4400\,\mathop {\text{A}}\limits^ \circ \cr} $$
35.
An unpolarised beam of intensity $${I_0}$$ is incident on a pair of nicols making an angle of $${60^ \circ }$$ with each other. The intensity of light emerging from the pair is
According to Malus’ law
$$I = {I_0}{\cos ^2}\theta = {I_0}\left( {{{\cos }^2}{{60}^ \circ }} \right) = {I_0} \times {\left( {\frac{1}{2}} \right)^2} = \frac{{{I_0}}}{4}.$$
36.
Consider Fraunh offer diffraction pattern obtained with a single slit illuminated at normal incidence. At the angular position of the first diffraction minimum the phase difference (in radians) between the wavelets from the opposite edges of the slit is
Path difference between the opposite edges is $$\lambda .$$
For a phase difference of $$2\,\pi $$ we get a path diff. of $$\lambda .$$
37.
The intensity of a point source of light, $$S,$$ placed at a distance $$d$$ in front of a screen $$A,$$ is $${I_0}$$ at the center of the screen. Find the light intensity at the center of the screen if a completely reflecting plane mirror $$M$$ is placed at a distance $$d$$ behind the source, as shown in figure.
The width of the diffraction band is given by
$$\eqalign{
& \beta = \frac{{\lambda D}}{d} \cr
& \Rightarrow \beta \propto \lambda \cr
& \beta \propto D\,\,{\text{and}}\,\,\beta \propto \frac{1}{d} \cr} $$
40.
Two sources $${S_1}$$ and $${S_2}$$ emitting coherent light waves of wavelength $$\lambda $$ in the same phase are situated as shown. The distance $$OP,$$ so that the light intensity detected at $$P$$ is equal to that at $$O$$ is
Referring to the figure, the path difference between the two waves starting from $${S_1}$$ and $${S_2}$$ tums out to be $$\left( {2\lambda \cos \theta } \right) = n\lambda $$ where $$n$$ is taken as $$1$$ to get the point of maximum intensity which is the same as a point $$O.$$
Therefore, the above relation gives $$\cos \theta = \frac{1}{2}$$ so that $$\theta = {60^ \circ }$$ and $$\tan \theta = \frac{{PO}}{D} = \sqrt 3 ,$$ giving $$PO = D\sqrt 3 .$$