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

The image $$I'$$ for first refraction (i.e., when the ray comes out of liquid) is at a depth of
$$ = \frac{{33.25}}{{1.33}} = 25\,cm\,\,\left[ {\because \,\,{\text{Apparent depth = }}\frac{{{\text{Real depth}}}}{\mu }} \right]$$
Now, reflection will occur at concave mirror. For this $$I'$$ behaves as an object.
$$\eqalign{ & \therefore \,\,u = - \left( {15 + 25} \right) = - 40\,cm \cr & {\text{and }}v = - \left[ {15 + \frac{{25}}{{1.33}}} \right] \cr} $$
Where $${\frac{{25}}{{1.33}}}$$  is the real depth of the image.
Using mirror formula we get
$$\eqalign{ & \frac{1}{f} = \frac{1}{v} + \frac{1}{u}, \cr & f = - 18.31\,cm \cr} $$

Gamma rays have wavelength range $$6 \times {10^{ - 14}}m$$   to $$1 \times {10^{ - 10}}m,$$   X-rays have wavelength range $$1 \times {10^{ - 13}}m$$   to $$3 \times {10^{ - 8}}m.$$   Blue light and red. light lies in visible range of spectrum which extends from $$4000\,\mathop {\text{A}}\limits^ \circ $$  to $$7800\,\mathop {\text{A}}\limits^ \circ .$$  Hence, wavelength of red light is longest.

For first minima at $$P,\,a\sin \theta = n\lambda $$

where, $$N = 1$$
$$ \Rightarrow a\sin \theta = \lambda $$
So, phase difference,
$$\eqalign{ & \Delta {\phi _1} = \frac{{\Delta {x_1}}}{\lambda } \times 2\pi = \frac{{\left( {\frac{a}{2}} \right)\sin \theta }}{\lambda } \times 2\pi \cr & = - \frac{\lambda }{{2\lambda }} \times 2\pi = \pi \,rad \cr} $$

$$\eqalign{ & {\text{Given,}}\,\mu = \frac{4}{3}h = 15\,cm \cr & \frac{{\sin {{90}^ \circ }}}{{\sin C}} = \mu \Rightarrow \sin C = \frac{1}{\mu } = \frac{R}{{\sqrt {{R^2} + {h^2}} }} = \frac{3}{4} \cr & \Rightarrow 16\,{R^2} = 9\,{R^2} + 9\,{h^2} \cr & {\text{or,}}\,\,7\,{R^2} = 9\,{h^2} \cr & {\text{or,}}\,R = \frac{3}{{\sqrt 7 }}h = \frac{3}{{\sqrt 7 }} \times 15\,cm \cr} $$

The ray diagram for the problem is shown as follows

According to lens formula, $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$
We have $$u = - 30\,cm,f = 20\,cm$$
$$\eqalign{ & \therefore \frac{1}{{20}} = \frac{1}{v} - \frac{1}{{ - 30}} \cr & {\text{or}}\,\,\frac{1}{v} = \frac{1}{{20}} - \frac{1}{{30}} = \frac{{3 - 2}}{{60}} = \frac{1}{{60}} \cr & \therefore v = 60\,cm \cr} $$
For the image $$\left( I \right)$$  coincident with object $$\left( O \right),$$  the rays after refraction from the lens must fall on the convex mirror normally or the rays refracted from lens must meet at $$C.$$
$$\therefore LC = v = 60\,cm$$
Thus, distance between lens and mirror
$$LM = 60 - 10 = 50\,cm$$

Angular resolution of telescope is given by $$ = \frac{{1.22\lambda }}{d}$$
$$\eqalign{ & = \frac{{1.22 \times 5000 \times {{10}^{ - 10}}}}{{10 \times {{10}^{ - 2}}}} \cr & = 6.1 \times {10^{ - 6}} \cr & \approx {10^{ - 6}}\,rad \cr} $$

We know that $$i + e - A = \delta $$
$$\eqalign{ & {35^ \circ } + {79^ \circ } - A = {40^ \circ } \cr & \therefore \,\,A = {74^ \circ } \cr} $$
$$\eqalign{ & {\text{But, }}\mu = \frac{{\sin \left( {\frac{{A + {\delta _m}}}{2}} \right)}}{{\frac{{\sin A}}{2}}} \cr & = \frac{{\sin \left( {\frac{{74 + {\delta _m}}}{2}} \right)}}{{\sin \frac{{74}}{2}}} \cr & = \frac{5}{3}\sin \left( {{{37}^ \circ } + \frac{{{\delta _m}}}{2}} \right) \cr} $$
$${\mu _{\max }}$$ can be $$\frac{5}{3}.$$ That is $${\mu _{\max }}$$ is less than $$\frac{5}{3}$$ $$ = 1.67$$
But $${{\delta _m}}$$ will be less than 40° so
$$\eqalign{ & \mu < \frac{5}{3}\sin {57^ \circ } < \frac{5}{3}\sin {60^ \circ } \cr & \Rightarrow \,\,\mu = 1.45 \cr} $$

NOTE : The intermediate image in compound microscope is real, inverted and magnified.

As shown in the figure, when the object $$(A)$$  is placed between $$F$$ and $$C,$$ the image $$(I)$$  is formed beyond $$C.$$ It is in this condition that when the student shifts his eyes towards left, the image appears to the right of the object pin. (Image distance > object distance)

If biconvex lens behaves like a plane sheet of glass, ray will pass undeviated through it only when medium has same refractive index as that of biconvex lens.