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

In this case, one of the image will be real and the other virtual. Let us assume that image of $${S_1}$$ is real and that of $${S_2}$$ is virtual.

Applying $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
For $${S_1}:\frac{1}{y} + \frac{1}{x} = \frac{1}{9}\,......\left( {\text{i}} \right)$$
For $${S_2}:\frac{1}{y} + \frac{1}{{24 - x}} = \frac{1}{9}\,......\left( {{\text{ii}}} \right)$$
Solving eqs. (i) and (ii),
we get $$x = 6\,cm$$

An optical fibre is a device based on total internal reflection by which a light signal can be transferred from one place to the other with a negligible loss of energy.
It consists of a very long and thin fibre of quartz glass.

When a light ray is incident at one end $$A$$ of fibre making a small angle of incidence. It suffers multiple total internal reflections and finally it reaches the point $$B.$$

Consider the activity $$A$$ to $$B$$
Applying $${v^2} - {u^2} = 2\,as$$
$$\eqalign{ & {v^2} - {0^2} = 2 \times 10 \times 7.2 \cr & \Rightarrow \,\,v = 12\,m/s \cr} $$
The velocity of ball as perceived by fish is
$$\eqalign{ & v' = {\,_w}\mu \times v \cr & = \frac{4}{3} \times 12 \cr & = 16\,m/s \cr} $$

For critical angle $${\theta _c},$$
$$\sin {\theta _c} = \frac{1}{\mu }$$
For greater wavelength or lesser frequency $$\mu $$ is less.
So, critical angle would be more. So, they will not suffer reflection and come out at angles less then 90°.

When refracting angle of a prism is small $$\left( {\angle {{10}^ \circ }} \right),$$   the deviation $$\delta $$ is calculated from the relation $$\delta = \left( {\mu - 1} \right)A.$$    For prisms with bigger refracting angles, we use the relation
$$\delta = \left( {{i_1} + {i_2}} \right) - A$$

Let thickness of the given slab is $$t.$$ According to the question, when viewed from both the surfaces

$$\eqalign{ & \frac{x}{\mu } + \frac{{t - x}}{\mu } = 3 + 5 \cr & \Rightarrow \frac{t}{\mu } = 8\,cm \cr & \Rightarrow t = 8 \times \mu \cr & \Rightarrow t = 8 \times \frac{3}{2} = 12\,cm \cr} $$

As shown in figure, a light ray from the coin will not emerge out of liquid, if $$i > C.$$

Therefore, minimum radius $$R$$ corresponds to $$i = C.$$

In $$\Delta SAB,$$
$$\eqalign{ & \frac{R}{h} = \tan C \cr & {\text{or}}\,\,R = h\tan C \cr & {\text{or}}\,\,R = \frac{h}{{\sqrt {{\mu ^2} - 1} }} \cr & {\text{Given,}}\,\,R = 3\,cm,h = 4\,cm \cr & {\text{Hence,}}\,\,\frac{3}{4} = \frac{1}{{\sqrt {{\mu ^2} - 1} }} \cr & {\text{or}}\,\,{\mu ^2} = \frac{{25}}{9}\,\,{\text{or }}\mu = \frac{5}{3} \cr & {\text{But}}\,\,\mu = \frac{c}{v}\,\,{\text{or}}\,\,v = \frac{c}{\mu } \cr & = \frac{{3 \times {{10}^8}}}{{\frac{5}{3}}} \cr & = 1.8 \times {10^8}\,m/s \cr} $$

This dark ring will be visible if ray from source gets total internal reflection from the spherical shell.

Let the source at any instant be at point $$P$$ then at point $$Q$$ ray will be totally reflected if $$\theta $$ is equal to or greater than critical angle. If $$QP$$  is equal to $$x,$$ then
$$z = \cos \theta = \frac{{{R^2} + {x^2} - {v^2}{t^2}}}{{2Rx}}$$
For $$\theta $$ to be minimum
$$\eqalign{ & \frac{{dz}}{{dx}} = \frac{{2x\left( {2Rx} \right) - 2R\left( {{R^2} + {x^2} - {v^2}{t^2}} \right)}}{{4{R^2}{x^2}}} = 0 \cr & \Rightarrow x = \sqrt {{R^2} - {v^2}{t^2}} \cr & {\text{So,}}\,\cos \theta = \frac{{2\left( {{R^2} - {v^2}{t^2}} \right)}}{{2R\sqrt {{R^2} - {v^2}{t^2}} }} = \frac{{\sqrt {{R^2} - {v^2}{t^2}} }}{R} \cr} $$
For no light come out, $$\sin \theta \geqslant \frac{1}{{\sqrt 2 }}\,\,{\text{or}}\,\,\theta \geqslant {45^ \circ }$$
$$\eqalign{ & \frac{{\sqrt {{R^2} - {v^2}{t^2}} }}{R} = \frac{1}{{\sqrt 2 }}; \cr & t = \frac{R}{{\sqrt 2 V}} \cr} $$

1^st minima is formed at a distance $$Y = \frac{{\lambda D}}{a}$$
For the distance of the first dark band of the diffraction pattern from the centre of the screen is given by position of 1^st minima.
i.e. $$Y = \frac{{\lambda D}}{a}$$
where, $$\lambda =$$  wavelength of parallel beams
$$D =$$  focal length
$$a =$$  width of linear aperture.
$$\eqalign{ & \Rightarrow Y = \frac{{\left( {5 \times {{10}^{ - 5}}} \right)\left( {0.6} \right)}}{{0.02 \times {{10}^{ - 2}}}}\,\,\left( {{\text{given}}} \right) \cr & \Rightarrow Y = 0.15\,cm \cr} $$

For total internal reflection to take place,
angle of incidence > critical angle
i.e. $$\theta > C$$
or $$\sin \theta > \sin C$$
But for the case of total internal reflection,
$$\sin C = \frac{1}{\mu }$$
and according to question
$$\eqalign{ & \theta = {90^ \circ } - r \cr & {\text{So,}}\,\,\sin \left( {{{90}^ \circ } - r} \right) > \frac{1}{\mu } \cr & {\text{i}}{\text{.e}}{\text{.}}\,\mu > \frac{1}{{\cos r}}\,\,\left[ {\therefore \sin \left( {{{90}^ \circ } - r} \right) = \cos r} \right]\,......\left( {\text{i}} \right) \cr} $$
From Snell's law,
$$\eqalign{ & \frac{{\sin {{45}^ \circ }}}{{\sin r}} = \mu \cr & \Rightarrow \sin r = \frac{1}{{\sqrt 2 \mu }} \cr & \therefore \cos r = \sqrt {1 - {{\sin }^2}r} \cr & = \sqrt {1 - \frac{1}{{2{\mu ^2}}}} \cr} $$
Thus, Eq. (i) becomes
$$\eqalign{ & \mu > \frac{1}{{\sqrt {1 - \frac{1}{{2{\mu ^2}}}} }} \cr & \therefore {\mu ^2} = \frac{1}{{1 - \frac{1}{{2{\mu ^2}}}}} \cr & {\text{or}}\,\,{\mu ^2} - \frac{1}{2} = 1\,\,{\text{or}}\,\,\mu = \sqrt {\frac{3}{2}} \cr} $$