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

Focal length of the combination
$$\frac{1}{f} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}}\,......\left( {\text{i}} \right)$$
We have $$\frac{1}{{{f_1}}} = \left( {{\mu _1} - 1} \right)\left( {\frac{1}{\infty } - \frac{1}{{ - R}}} \right) = \frac{{{\mu _1} - 1}}{R}$$
and $$\frac{1}{{{f_2}}} = \left( {{\mu _2} - 1} \right)\left( {\frac{1}{{ - R}} - \frac{1}{\infty }} \right) = - \frac{{\left( {{\mu _2} - 1} \right)}}{R}$$
Putting these values of $$\frac{1}{{{f_1}}}$$ and $$\frac{1}{{{f_2}}}$$ in Eq. (i)
$$\eqalign{ & \frac{1}{f} = \frac{{\left( {{\mu _1} - 1} \right)}}{R} - \frac{{\left( {{\mu _2} - 1} \right)}}{R} \cr & = \frac{{\left[ {{\mu _1} - 1 - {\mu _2} + 1} \right]}}{R} = \frac{{{\mu _1} - {\mu _2}}}{R} \cr & f = \frac{R}{{{\mu _1} - {\mu _2}}} \cr} $$

When a ray falls on convex surface of a planoconvex lens, then it is first refracted and reflected from plane surface and then finally refracted from convex surface. Thus, two refractions and one reflection take place.
Since, refraction takes place two times and reflection takes place one time
So, focal length of planoconvex lens is
$$\frac{1}{F} = \frac{2}{{{f_l}}} + \frac{1}{{{f_m}}}\,......\left( {\text{i}} \right)$$
If plane surface is silvered, so
$${f_m} = \frac{{{R_2}}}{2} = \frac{\infty }{2} = \infty $$
And from lens maker formula
$$\eqalign{ & \frac{1}{{{f_l}}} = \left( {\mu - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right) \cr & = \left( {\mu - 1} \right)\left( {\frac{1}{R} - \frac{1}{\infty }} \right) = \frac{{\left( {\mu - 1} \right)}}{R} \cr & \therefore \frac{1}{F} = \frac{{2\left( {\mu - 1} \right)}}{R} + \frac{1}{\infty } \cr & = \frac{{2\left( {\mu - 1} \right)}}{R} \cr & {\text{or}}\,\,F = \frac{R}{{2\left( {\mu - 1} \right)}}\,......\left( {{\text{ii}}} \right) \cr & {\text{Given,}}\,\,R = 20\;cm,\mu = 1.5 \cr & {\text{Hence,}}\,\,F = \frac{{20}}{{2\left( {1.5 - 1} \right)}} \cr & = \frac{{20}}{{2 \times 0.5}} \cr & = 20\,cm \cr} $$

$$\frac{1}{f} = \left( {\frac{{{\mu _2}}}{{{\mu _1}}} - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)$$

If $${\mu _2} > {\mu _1},$$   the concave lens maintains its nature otherwise the nature of the lens will be reversed.
So, the lens should be filled with $${L_2}$$ and immerse in $${L_1}.$$

Power of a lens is the ability of the lens to converge a beam of light falling on the lens. It is measured as the reciprocal of focal length of the lens.
i.e. $$P = \frac{1}{{{f_{\left( {{\text{in}}\,m} \right)}}}}$$
When $$f$$ is in $$cm,$$
$$P = \frac{{100}}{f}D$$
Here, $${f_1} = 80\,cm,{f_2} = - 50\,cm$$
Combined power of two lenses in contact is given by $${P_{{\text{eq}}}} = {P_1} + {P_2}$$
$$\eqalign{ & \Rightarrow {P_{{\text{eq}}}} = \frac{1}{{{f_1}\left( m \right)}} + \frac{1}{{{f_2}\left( m \right)}}\,\,\left[ {{f_1}\,{\text{and}}\,{f_2}\,{\text{are with sign}}} \right] \cr & \therefore P = \frac{{100}}{{80}} - \frac{{100}}{{50}} \cr & = - 0.75\,D \cr} $$

Since the water has a greater index of refraction than the lens, the magnifier now acts like a diverging lens, resulting in an image that is smaller than the object.

The phenomenon of total internal reflection takes place during reflection at $$P.$$
$$\sin \theta = \frac{1}{{_g^w\mu }}\,\,\,\,.....\left( {\text{i}} \right)$$

Now, $$_g^w\mu = \frac{{_g^a\mu }}{{_w^a\mu }}$$
$$\eqalign{ & = \frac{{1.5}}{{\frac{4}{3}}} \cr & = 1.125 \cr & \therefore \,\,\sin \theta = \frac{1}{{1.125}} \cr & = \frac{8}{9} \cr} $$
$$\therefore \,\,\sin \theta $$   should be greater than $$\frac{8}{9}.$$

$$\eqalign{ & \therefore \,\,n = \frac{{{\text{Velocity of light in vacuum}}}}{{{\text{Velocity of light in medium}}}} \cr & \therefore \,\,n = \frac{3}{2} \cr & {3^2} + {\left( {R - 3\,mm} \right)^2} = {R^2} \cr & \Rightarrow \,\,{3^2} + {R^2} - 2\,R\left( {3\,mm} \right) + {\left( {3\,mm} \right)^2} = {R^2} \cr & \Rightarrow \,\,R \approx 15\,cm \cr & \frac{1}{f} = \left( {\frac{3}{2} - 1} \right)\left( {\frac{1}{{15}}} \right) \cr & \Rightarrow \,\,f = 30\,cm \cr} $$

Since both surfaces have same radius of curvature on the same side, no dispersion will occur.

Given, $${Y_{{\text{steel}}}} = 2{Y_{{\text{brass}}}}\,{\text{and}}\,{L_s} = {L_b}\,{\text{and}}\,{A_s} = {A_b},{\text{such}}\,{\text{that}}\,\Delta {L_s} = \Delta {L_b}$$
As we know, Young's modulus,
$$Y = \frac{{{\text{stress}}}}{{{\text{strain}}}} = \frac{{\frac{F}{A}}}{{\frac{{\Delta L}}{L}}} = \frac{{W\Delta L}}{{A\Delta L}}$$
$$\eqalign{ & {\text{So,}}\,\,W = \frac{{FA\Delta L}}{L} \propto Y \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,\frac{{{W_s}}}{{{W_b}}} = \frac{{{Y_s}}}{{{Y_b}}} = \frac{{2{Y_b}}}{{{Y_b}}} = \frac{2}{1} \cr & \Rightarrow 2:1 \cr} $$
Thus, weight added to the steel and brass wires must be in the ratio of $$2:1.$$

Use lens maker's formula
$$\eqalign{ & \frac{1}{f} = \left( {_g^m\mu - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right) \cr & {\text{Now,}}\,\,_g^m\mu = \frac{{_g\mu }}{{_m\mu }} = \frac{{1.5}}{{1.75}} \cr & \therefore \frac{1}{f} = \left( {\frac{{1.5}}{{1.75}} - 1} \right)\left( { - \frac{1}{R} - \frac{1}{R}} \right) = + \frac{{0.25 \times 2}}{{1.75\,R}} \cr & \Rightarrow f = + 3.5\,R \cr} $$
The positive sign shows that the lens behaves as convergent lens.