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

Let $$x$$ be thickness of window, $$v$$ be velocity of light entering in window, so time taken by sunlight to pass through the window.
$$\eqalign{ & t = \frac{x}{v} = \frac{x}{{\frac{c}{\mu }}} = \frac{{\mu x}}{c}\,\,\left[ {v = \frac{c}{\mu }} \right] \cr & \therefore t = \frac{{1.5 \times 4 \times {{10}^{ - 3}}}}{{3 \times {{10}^8}}} \cr & = 2 \times {10^{ - 11}}s \cr} $$

For T.I.R. $${45^ \circ } < C$$
$$\eqalign{ & \therefore \sin {45^ \circ } < \sin C \cr & \therefore \frac{1}{{\sqrt 2 }} > \frac{{\frac{4}{3}}}{n} \cr & \therefore n > \frac{{4\sqrt 2 }}{3} \cr} $$

When white light is incident on a soap bubble it is partly reflected from upper surface and partly reflected from lower surface. These two reflected beams superpose to cause interference. The colours which satisfy the condition of maxima are visible in reflected light. So, colours of soap bubbles are caused due to interference.

Given,
Magnification $$ = \frac{{{f_o}}}{{{f_e}}} = 9$$   and $${f_o} + {f_e} = 20$$
$${f_o} = 9{f_e}$$
$$\eqalign{ & {\text{So,}}\,\,9{f_e} + {f_e} = 20 \cr & {f_e} = 2\,cm \cr & \therefore {f_o} = 9 \times 2 \cr & {f_o} = 18\,cm \cr} $$

For equal size of image and object, object must be placed of centre of curvature of lens i.e. $$u = v$$
The lens formula can be written as
$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\,......\left( {\text{i}} \right)$$
Given, $$v = d$$
For equal sized image,
$$\left| v \right| = \left| u \right| = d$$
By sign convention,
$$\eqalign{ & u = - d \cr & \therefore \frac{1}{f} = \frac{1}{d} + \frac{1}{d}\,\,{\text{or}}\,\,f = \frac{d}{2} \cr} $$

Given, Focal length of objective, $${f_0} = 30\,cm$$
focal length of eye lens, $${f_e} = 3.0\,cm$$
Magnifying power, $$M = ?$$
Magnifying power of the Galilean telescope,
$$\eqalign{ & {M_D} = \frac{{{f_0}}}{{{f_e}}}\left( {1 - \frac{{{f_e}}}{D}} \right) = \frac{{30}}{3}\left( {1 - \frac{3}{{25}}} \right)\,\,\left[ {\because D = 25\,cm} \right] \cr & = 10 \times \frac{{22}}{{25}} = 8.8\,cm \cr} $$

The minimum size of plane mirror required to see full length of a man $$ = \frac{{{\text{height of man}}}}{2}$$
Given, height of man $$= 6\,ft$$
Thus, minimum size of plane mirror $$ = \frac{6}{2} = 3\,ft$$

As the object and image distance is same, object is placed at $$2\,f,$$  Therefore $$2\,f = 10$$
or $$f = 5\,cm.$$
Shift due to slab, $$d = t\left( {1 - \frac{1}{\mu }} \right)$$
in the direction of incident ray
$$ \Rightarrow \,\,d = 1.5\left( {1 - \frac{2}{3}} \right) = 0.5\,cm$$
Now, $$u = - 9.5\,cm$$
Again using lens formulas $$\frac{1}{v} - \frac{1}{{ - 9.5}} = \frac{1}{5}$$
⇒ $$v = 10.55\,cm$$
Thus, screen is shifted by a distance $$d = 10.55 - 10 = 0.55\,cm\,$$     away from the lens.

We will have single surface refractions successively at the four surfaces $${S_1},{S_2},{S_3}$$   and $${S_4}.$$ Do not forget to shift origin to the vertex of respective surface.
Refractive at first surface $${S_1}$$ : Light travels from air to glass.
$$\eqalign{ & \frac{{1.5}}{{{v_1}}} - \frac{1}{\infty } = \frac{{1 - 1.5}}{{\left( { + 10} \right)}} \cr & {v_1} = 30\,cm \cr} $$
First image is object for the refraction at second surface. For refraction at surface $${S_2}$$ : Light travels from glass to air
$$\eqalign{ & \frac{1}{{{v_2}}} - \frac{{1.5}}{{\left( { + 25} \right)}} = \frac{{1 - 1.5}}{{\left( + \right)5}} \cr & {v_2} = - 25\,cm \cr} $$
For refraction at surface $${S_3}$$ : Light travels from air to glass.
$$\eqalign{ & \frac{{1.5}}{{{v_3}}} - \frac{1}{{\left( { - 35} \right)}} = \frac{{\left( {1.5 - 1} \right)}}{{\left( { - 5} \right)}} \cr & {v_3} = \frac{{ - 35}}{3}cm \cr} $$
For refraction at surface $${S_4}$$ : Light travels from glass to air
similarly, $${v_4} = - 25\,cm$$
The final image is virtual, formal at $$25\,cm$$  to the left of the vertex of surface $${S_4}.$$

For net intensity
$$I' = 4{I_0}{\cos ^2}\frac{\phi }{2}$$
For the first case, $$K = 4{I_0}{\cos ^2}\left[ \pi \right]\because \left( {\phi = \frac{{2\pi }}{\lambda } \times \lambda } \right)$$
$$K = 4{I_0}\,......\left( \operatorname{i} \right)$$
For the second case,
$$\eqalign{ & K' = 4{I_0}{\cos ^2}\left( {\frac{{\frac{\pi }{2}}}{2}} \right) \cr & \because \left( {\phi = \frac{{2\pi }}{\lambda } \times \frac{\lambda }{4}} \right) \cr & K' = 2{I_0}\,......\left( {{\text{ii}}} \right) \cr} $$
Comparing Eqs. (i) and (ii), we get
$$K' = \frac{K}{2}$$