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

As, resolving power of a microscope,
$$\eqalign{ & \left( {RP} \right) \propto \frac{1}{{\lambda \left( {{\text{wavelength}}} \right)}} \cr & \therefore \frac{{R{P_1}}}{{R{P_2}}} = \frac{{{\lambda _2}}}{{{\lambda _1}}} = \frac{{6000}}{{4000}} = \frac{3}{2} \cr} $$

Convex lens forms the image at $${I_1}.$$ $${I_1}$$ is at the second focus of convex lens. Size of $${I_1}$$ $$= 2cm.$$
$${I_1}$$ acts as virtual object for concave lens. Concave lens forms the image of $${I_1}$$ and $${I_2}.$$

For concave lens,
$$\eqalign{ & \frac{1}{v} - \frac{1}{4} = - \frac{1}{{20}}\,\,{\text{or }}\frac{1}{v} = - \frac{1}{{20}} + \frac{1}{4} \cr & = \frac{4}{{20}} \cr & = \frac{1}{5} \cr} $$
or $$v$$ = $$5cm$$  = Distance of $${I_2}$$ from concave lens.
∴ Magnification $$ = \frac{v}{u} = \frac{{{\text{size of image}}}}{{{\text{size of object}}}} = \frac{5}{4}$$
or $$\frac{{{\text{size of image}}}}{2} = 1.25$$
or size of image due to concave lens = $$2.5\,cm$$

NOTE :
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.
$$\left( {\because \nu = \nu \lambda } \right)$$

From mirror formula
$$\eqalign{ & \frac{1}{v} + \frac{1}{u} = \frac{1}{f}\,\,{\text{so,}}\frac{{dv}}{{dt}} = \frac{{{v^2}}}{{{u^2}}}\left( {\frac{{du}}{{dt}}} \right) \cr & \Rightarrow \,\,\frac{{dv}}{{dt}} = - {\left( {\frac{f}{{u - f}}} \right)^2}\frac{{du}}{{dt}} \cr & \Rightarrow \,\,\frac{{dv}}{{dt}} = \frac{1}{{15}}m/s \cr} $$

Focal length of convex lens in air is given by
$$\frac{1}{{{f_a}}} = \left( {_a{\mu _g} - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)\,......\left( {\text{i}} \right)$$
Focal length of convex lens when immersed in liquid of refractive index $${\mu _l}$$ is given by
$$\frac{1}{{{f_l}}} = \left( {_l{\mu _g} - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)$$
$${_l{\mu _g}} = $$   refractive index of glass w.r.t. liquid
$${R_1}{R_2} = $$   radius of curvature of lens spherical surface
$${\text{or}}\,\,\frac{1}{{{f_l}}} = \left( {\frac{{{\mu _g}}}{{{\mu _l}}} - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)\,......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii),
$$\eqalign{ & \frac{{\frac{1}{{{f_a}}}}}{{\frac{1}{{{f_l}}}}} = \frac{{\left( {{\mu _g} - 1} \right)}}{{\left( {\frac{{{\mu _g}}}{{{\mu _l}}} - 1} \right)}} \cr & {\text{or}}\,\,\frac{{{f_l}}}{{{f_a}}} = \frac{{\left( {{\mu _g} - 1} \right)}}{{\left( {\frac{{{\mu _g}}}{{{\mu _l}}} - 1} \right)}} \cr & {\text{or}}\,\,\frac{{{f_l}}}{{{f_a}}} = \frac{{\left( {1.5 - 1} \right)}}{{\left( {\frac{{1.5}}{{1.25}} - 1} \right)}} \cr & = \frac{{\frac{1}{2}}}{{\frac{1}{5}}} = \frac{5}{2} \cr & \therefore {f_l} = \frac{5}{2}{f_a} = \frac{5}{2} \times 2 = 5\,cm \cr} $$

From similar triangles,
$$\eqalign{ & \frac{{QC}}{{\sin {{30}^ \circ }}} = \frac{R}{{\sin {{120}^ \circ }}} \cr & {\text{or}}\,\,QC = R \times \frac{{\sin {{30}^ \circ }}}{{\sin {{120}^ \circ }}} = \frac{R}{{\sqrt 3 }} \cr} $$
Thus $$PQ = PC - QC = R - \frac{R}{{\sqrt 3 }} = R\left( {1 - \frac{1}{{\sqrt 3 }}} \right)$$

We resolve the velocity vector $${\vec v}$$ of the person into two components, one parallel to the mirror, $${{\vec v}_{\left\| {} \right.}}$$ and the other perpendicular to the mirror, $${{\vec v}_ \bot },$$  i.e. $$\vec v = {{\vec v}_{\left\| {} \right.}} + {{\vec v}_ \bot }$$  (figure).
The velocity of the image will obviously be $$\vec v' = {{\vec v}_{\left\| {} \right.}} + {{\vec v}_ \bot }.$$   Therefore, the velocity at which the person approaches his image is defined as his velocity relative to the image. From the formula $${v_{{\text{rel}}}} = 2{{\vec v}_ \bot } = 2v\sin \alpha .$$

$$\eqalign{ & \frac{1}{f} = \left( {\mu - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right); \cr & R = 10\,cm. \cr & \frac{1}{{f'}} = \left( {3 - 1} \right)\left( {\frac{1}{{ - 10}} - \frac{1}{{10}}} \right), \cr & f' = - \frac{{10}}{4} \cr & \frac{1}{{{f_{{\text{eq}}}}}} = \frac{1}{{20}} - \frac{4}{{10}} + \frac{1}{{20}} = \frac{2}{{20}} - \frac{4}{{10}} \Rightarrow {f_{{\text{eq}}}} = - \frac{{10}}{3} \cr} $$

For lens $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\eqalign{ & \frac{1}{v} - \frac{1}{{ - 50}} = \frac{1}{{30}} \cr & \therefore \,\,v = 75\,cm \cr} $$

Therefore object distance for mirror is $$25cm$$  and object is virtual
For mirror $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
$$\eqalign{ & \therefore \,\,\frac{1}{v} + \frac{1}{{25}} = \frac{1}{{50}} \cr & \therefore \,\,v = - 50\,cm \cr} $$
The image $$I$$ would have formed as shown had the mirror been straight. But here the mirror is tilted by 30°.
Therefore the image will be tilted by 60° and will be formed at $$A.$$
Here $$MA = 50\,\cos {60^ \circ } = 25\,cm$$
and $$I'A = 50\,\sin {60^ \circ } = 25\sqrt 3\, cm$$

According to question, the ray diagram will be

So, distance $$d$$ between convex lens and concave mirror is given by
$$d = 2{f_1} + {f_2}$$