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

Optical fibres form a dielectric wave guide and are free from electromagnetic interference or radio frequency interference.

$$\eqalign{ & A = {90^ \circ } - \theta \cr & \Rightarrow {r_2} = A = {90^ \circ } - \theta > {\theta _c} \cr & \cos \theta > \sin {\theta _c} = \frac{{\frac{6}{5}}}{{\frac{2}{3}}} = \frac{4}{5}\,\,\left( {{\theta _c}\,{\text{is critical angle}}} \right) \cr & \theta < {\cos ^{ - 1}}\frac{4}{5} = {37^ \circ } \cr} $$

$$\eqalign{ & {\text{For}}\,P:\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \cr & {\text{or}}\,\,\frac{1}{{{v_P}}} + \frac{1}{{ - 3}} = \frac{1}{{ - 1}} \cr & {\text{or}}\,\,{v_P} = - \frac{3}{2}m \cr & {\text{For}}\,Q:\frac{1}{{{v_Q}}} + \frac{1}{{ - 5}} = \frac{1}{{ - 1}} \cr & {\text{or}}\,\,{v_Q} = \frac{{ - 5}}{4}m \cr} $$
So horizontal distance between image of $$P$$ and $$Q$$ is $$0.25\,m$$

For minimum deviation the ray in the prism is parallel to the base of the prism. This condition does not depend on the colour (or wave length) of incident radiation. So in both the cases, by geometry, $$r = {30^ \circ }.$$   So (A) is correct option.

$$\eqalign{ & \cos \left( {{{180}^ \circ } - 2\alpha } \right) = \frac{{\left( {\frac{{\hat i}}{2} + \frac{{\sqrt 3 }}{2}\hat j} \right).\left( {\frac{{\hat i}}{2} - \frac{{\sqrt 3 }}{2}\hat j} \right)}}{{\sqrt {{{\left( {\frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \sqrt {\left( {\frac{1}{2}} \right) + {{\left( { - \frac{{\sqrt 3 }}{2}} \right)}^2}} }} \cr & \therefore \,\,\cos \left( {{{180}^ \circ } - 2\alpha } \right) = - \frac{1}{2} \cr & \therefore \,\,{180^ \circ } - 2\alpha = {120^ \circ } \cr & \therefore \,\,\alpha = {30^ \circ } \cr} $$
option (A) is correct

The focal length of the lens
$$\eqalign{ & \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \cr & = \frac{1}{{12}} + \frac{1}{{240}} \cr & = \frac{{20 + 1}}{{240}} \cr & = \frac{{21}}{{240}} \cr & f = \frac{{240}}{{21}}cm \cr} $$
Shift $$ = t\left( {1 - \frac{1}{\mu }} \right)$$
$$\eqalign{ & \Rightarrow \,\,1\left( {1 - \frac{1}{{\frac{3}{2}}}} \right) \cr & = 1 \times \frac{1}{3} \cr} $$
Now, $$v' = 12 - \frac{1}{3}$$
$$ = \frac{{35}}{3}cm$$
Now the object distance $$u.$$
$$\eqalign{ & \frac{1}{u} = \frac{3}{{35}} - \frac{{21}}{{240}} \cr & = \frac{1}{5}\left[ {\frac{3}{7} - \frac{{21}}{{48}}} \right] \cr & \frac{1}{u} = \frac{1}{5}\left[ {\frac{{48 - 49}}{{7 \times 16}}} \right] \cr & u = - 7 \times 16 \times 5 \cr & = - 560\,cm \cr & = - 5.6\,m \cr} $$

$$\eqalign{ & {\text{Given: }}{f_0} = 50\,cm,{f_e} = 5\,cm \cr & d = 25\,cm,{u_0} = - 200\,cm \cr & {\text{Magnification }}M = ? \cr & {\text{As }}\frac{1}{{{v_0}}} - \frac{1}{{{u_0}}} = \frac{1}{{{f_0}}} \cr & \Rightarrow \frac{1}{{{v_0}}} = \frac{1}{{{f_0}}} + \frac{1}{{{u_0}}} = \frac{1}{{50}} - \frac{1}{{200}} = \frac{{4 - 1}}{{200}} = \frac{3}{{200}} \cr & {\text{or }}{v_0} = \frac{{200}}{3}\,cm \cr & {\text{Now }}{v_e} = d = - 25\,cm \cr & {\text{From,}}\,\,\frac{1}{{{v_e}}} - \frac{1}{{{u_e}}} = \frac{1}{{{f_e}}} \cr & - \frac{1}{{{u_e}}} = \frac{1}{{{f_e}}} - \frac{1}{{{v_e}}} = \frac{1}{5} + \frac{1}{{25}} = \frac{6}{{25}} \cr & {\text{or, }}{v_e} = \frac{{ - 25}}{6}\,cm \cr & {\text{Magnification }}M = {M_0} \times {M_e} \cr & = \frac{{{v_0}}}{{{u_0}}} \times \frac{{{v_e}}}{{{u_e}}} = \frac{{\frac{{ - 200}}{3}}}{{200}} \times \frac{{ - 25}}{{\frac{{ - 25}}{6}}} = - \frac{1}{3} \times 6 = - 2 \cr} $$

Angle of prism is given by $$A = {r_1} + {r_2}$$
where, $${r_1}$$ is refraction angle on incident face and $${r_2}$$ is angle of incidence on 2nd face of prism. As refracted ray emerges normally from opposite surface, $${r_2} = 0.$$

$$\eqalign{ & \therefore A = {r_1} \cr & {\text{Now,}}\,\,\mu = \frac{{\sin i}}{{\sin {r_1}}} \cr} $$
If $$\angle {i_1}$$  and $$\angle {r_1}$$  are very small then, $$\sin i \approx i,\sin {r_1} \approx {r_1}$$
$$\eqalign{ & \therefore \mu = \frac{i}{{{r_1}}} = \frac{i}{A} \cr & \therefore i = \mu A \cr} $$

Maximum number of reflection $$ = \frac{{2\sqrt 3 }}{x}$$
where $$x = 0.2\tan {30^ \circ } = \frac{{0.2}}{{\sqrt 3 }}.$$

The distance of bottom of the beaker from mirror $$ = h - d\left( {1 - \frac{1}{\mu }} \right)$$
So it will be at a distance $$ = h - d\left( {1 - \frac{1}{\mu }} \right)$$    from mirror. Now distance between bottom of beaker and image
$$\eqalign{ & = h + h - d\left( {1 - \frac{1}{\mu }} \right) \cr & = 2h - d\left( {\frac{{\mu - 1}}{\mu }} \right). \cr} $$