131.
A ray of light is incident at the glass-water interface at an angle $$i,$$ it emerges finally parallel to the surface of water, then the value of $${\mu _g}$$ would be
133.
The interference pattern is obtained with two coherent light sources of intensity ratio $$n.$$ In the interference pattern, the ratio $$\frac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}}$$ will be
A
$$\frac{{\sqrt n }}{{n + 1}}$$
B
$$\frac{{2\sqrt n }}{{n + 1}}$$
C
$$\frac{{\sqrt n }}{{{{\left( {n + 1} \right)}^2}}}$$
D
$$\frac{{2\sqrt n }}{{{{\left( {n + 1} \right)}^2}}}$$
It is given that, $$\frac{{{I_2}}}{{{I_1}}} = n \Rightarrow {I_2} = n{I_1}$$
$$\therefore $$ Ratio of intensities is given by
$$\eqalign{
& \frac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}} = \frac{{{{\left( {\sqrt {{I_2}} + \sqrt {{I_1}} } \right)}^2} - {{\left( {\sqrt {{I_2}} - \sqrt {{I_1}} } \right)}^2}}}{{{{\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)}^2} + \left. {{{\left( {\sqrt {{I_2}} - \sqrt {{I_1}} } \right)}^2}} \right)}} \cr
& = \frac{{{{\left( {\sqrt {\frac{{{I_2}}}{{{I_1}}}} + 1} \right)}^2} - {{\left( {\sqrt {\frac{{{I_2}}}{{{I_1}}}} - 1} \right)}^2}}}{{{{\left( {\sqrt {\frac{{{I_2}}}{{{I_1}}}} + 1} \right)}^2} + {{\left( {\sqrt {\frac{{{I_2}}}{{{I_1}}}} - 1} \right)}^2}}} \cr
& = \frac{{{{\left( {\sqrt n + 1} \right)}^2} - {{\left( {\sqrt n - 1} \right)}^2}}}{{{{\left( {\sqrt n + 1} \right)}^2} + {{\left( {\sqrt n - 1} \right)}^2}}} \cr
& = \frac{{2\sqrt n }}{{n + 1}} \cr} $$
134.
Two plane mirrors $$A$$ and $$B$$ are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle 30° at a point just inside one end of $$A.$$ The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is
KEY CONCEPT : The resolving power of a telescope $$R.P = \frac{D}{{1.22\lambda }}$$ where $$D$$ = diameter of the objective lens
$$\lambda $$ = wave length of light
Clearly, larger the aperture, larger is the value of $$D,$$ more is the resolving power or resolution.
136.
The refracting angle of a prism is $$A,$$ and refractive index of the material of the prism is $$\cot \left( {\frac{A}{2}} \right).$$ The angle of minimum deviation is
137.
A transparent solid cylindrical rod has a refractive index of $${\frac{2}{{\sqrt 3 }}}.$$ It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure.
The incident angle $$\theta $$ for which the light ray grazes along the wall of the rod is:
A
$${\sin ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)$$
B
$${\sin ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right)$$
C
$${\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)$$
D
$${\sin ^{ - 1}}\left( {\frac{1}{{2}}} \right)$$
138.
A ray of light from a denser medium strike a rarer medium at an angle of incidence $$i$$ (see Fig). The reflected and retracted rays make an angle of $${{{90}^ \circ }}$$ with each other. The angles of reflection and refraction are $$r$$ and $$r’$$ The critical angle is
$$C = {\sin ^{ - 1}}\left( {\frac{1}{{_2^1\mu }}} \right)\,\,\,.....\left( {\text{i}} \right)$$
Applying Snell's law at $$P,$$ we get
$$\eqalign{
& _2^1\mu = \frac{{\sin r'}}{{\sin i}} \cr
& = \frac{{\sin \left( {90 - r} \right)}}{{\sin r}}\left[ {\because \,\,i = r,r' + r = {{90}^ \circ }} \right] \cr
& \because \,\,_2^1\mu = \frac{{\cos r}}{{\sin r}}\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
From (i) and (ii)
$$C = {\sin ^{ - 1}}\left( {\tan r} \right)$$
139.
Light travels through a glass plate of thickness $$t$$ and refractive index $$\mu .$$ If $$c$$ is the speed of light in vacuum, the time taken by light to travel this thickness of glass is
Here, total thickness $$= t$$
Refractive index $$ = \mu $$
Speed of light in glass plate $$ = \frac{c}{\mu }$$
$$\left[ {\because v = \frac{{{\text{speed of light in vacuum}}}}{{{\text{refractive index of medium}}}}} \right]$$
$$\therefore $$ Time taken by light to travel this thickness of glass
$$ = \frac{t}{{\left( {\frac{c}{\mu }} \right)}} = \frac{{\mu t}}{c}$$
140.
When a ray of light enters a glass slab from air,
A
its wavelength decreases.
B
its wavelength increases.
C
Its frequency decreases.
D
neither its wavelength nor its frequency changes.
When the ray enters a glass slab from air, its frequency remains unchanged.
Since glass slab in an optically denser medium, the velocity of light decreases and therefore we can conclude that the wavelength decreases.
$$\left( {\because \nu = \nu \lambda } \right)$$