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

The focal length $$f$$ of the equivalent mirror is
$$\eqalign{ & \frac{1}{f} = \frac{2}{{{f_1}}} + \frac{1}{{{f_m}}} \cr & = \frac{2}{{15}} + \frac{1}{\infty } \cr & \Rightarrow \,\,f = \frac{{15}}{2}cm, \cr } $$
Since $$f$$ has a positive value, the combination behaves as a converging mirror.
Here $$u = - 20\,cm,f = - \frac{{15}}{2}cm,$$
$$v = ?$$
According to mirror formula $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
$$\eqalign{ & \Rightarrow \,\,\frac{1}{v} - \frac{1}{{ - 20}} = \frac{1}{{ - \frac{{15}}{2}}} \cr & \Rightarrow \,\,v = - 12\,cm \cr} $$
Negative sign indicates that the image is $$12\,cm$$  in front of mirror.

In displacement method, total magnification $$m = \sqrt {{m_1}{m_2}} .$$    Therefore, area of source is given by $$A = \sqrt {{A_1}{A_2}} .$$

When two or more lenses are combined together in such a way that the combination is free from chromatic aberration, then such a combination is called achromatic combination of lenses. For this purpose, one lens should be convex and other a concave lens.

Real and apparent depth are explained on the basis of refraction only. The concept of TIR is not involved here.

Applying Snell's law at $$P,$$
$$^1{\mu _2} = \frac{{\sin i}}{{\sin {r_1}}} = \frac{{{\mu _2}}}{{{\mu _1}}}\,\,\,.....\left( 1 \right)$$

Applying Snell's law at $$Q,$$
$$^2{\mu _3} = \frac{{\sin {r_1}}}{{\sin {r_2}}} = \frac{{{\mu _3}}}{{{\mu _2}}}\,\,\,\,.....\left( 2 \right)$$
Again applying Snell's law at $$R$$
$$^3{\mu _4} = \frac{{\sin {r_2}}}{{\sin i}} = \frac{{{\mu _4}}}{{{\mu _3}}}\,\,\,\,.....\left( 3 \right)$$
Multiplying (i), (ii) and (iii), we get
$${\mu _4} = {\mu _1}$$
NOTE : If the emergent ray is parallel to incident ray after travelling a number of parallel interfaces then the refractive index of the first and the last medium is always same.

$${\mu _g}\sin i = {\mu _{{\text{air}}}}\sin {90^ \circ }{\mu _g} = \frac{1}{{\sin i}}$$

The axial magnification $$\delta v = - \frac{{{v^2}}}{{{u^2}}}\left( {\delta u} \right)$$
$$ = {\left( {\frac{f}{{u - f}}} \right)^2}\ell .$$

In the above figure, $${S_1}$$ and $${S_2}$$ are the two different slits.
Given, distance between slits $${S_1}$$ and $${S_2},$$ $$d = 5\,\lambda $$  distance between screen and slits, $$D = 10\,d = 50\,\lambda $$
Here, $$\lambda $$ is the wavelength of light used in the experiment.
According to question, the intensity at maximum in this Young's double slit experiment is $${I_0}.$$
$$ \Rightarrow {I_{\max }} = {I_0}$$
$$\because $$ Path difference $$ = \frac{{d{Y_n}}}{D} = \frac{{d \times \frac{d}{2}}}{{10\,d}} = \frac{d}{{20}} = \frac{\lambda }{4}\,\,\left\{ {\because d = 5\lambda } \right\}$$
A path difference of $$\lambda $$ corresponds to phase difference $$2\pi $$
So, for path difference $$\frac{\lambda }{4},$$  phase difference
$$\phi = \frac{{2\pi }}{\lambda } \times \frac{\lambda }{4} = \frac{\pi }{2} = {90^ \circ }$$
As we know, $$I = {I_0}{\cos ^2}\frac{\phi }{2}$$
$$\eqalign{ & \Rightarrow I = {I_0}{\cos ^2}\frac{{{{90}^ \circ }}}{2} \cr & \Rightarrow I = {I_0} \times {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \Rightarrow I = \frac{{{I_0}}}{2} \cr} $$

KEY CONCEPT : When two plane mirrors are inclined at each other at an angle $$\theta $$ then the number of the images of a point object placed between the plane mirrors is $$\frac{{{{360}^ \circ }}}{\theta } - 1,{\text{if }}\frac{{{{360}^ \circ }}}{\theta }$$     is even
∴ Number of images formed $$ = \frac{{{{360}^ \circ }}}{{{{60}^ \circ }}} - 1 = 5$$

Lens maker's formula is
$$\eqalign{ & \frac{1}{f} = \left( {\mu - 1} \right)\left( {\frac{1}{{{R_1}}} = \frac{1}{{{R_2}}}} \right) \cr & {\text{or}}\,\,\frac{1}{f} \propto \mu \,\,{\text{or}}\,\,f \propto \frac{1}{\mu }\,......\left( {\text{i}} \right) \cr} $$
According to Cauchy’s formula,
$$\mu \propto \frac{1}{\lambda }\,......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), we get
$$f \propto \lambda $$
Hence, focal length of a converging lens is maximum for red colour (highest wavelength) and minimum for violet colour (lowest wavelength)
$${\text{i}}{\text{.e}}{\text{.}}\,{f_V} < {f_R}$$