$$\eqalign{ & {\text{In}}\,\Delta ABD,\frac{{BD}}{{3H}} = \tan {45^ \circ } \cr & {\text{or}}\,BD = 3H \cr} $$

And in $$\Delta A'BC,\frac{{BC}}{{2H}} = \tan \theta = \frac{1}{2}$$
$${\text{or}}\,\,BC = H$$
Now, $$y = BD - BC = 2H - H = 2H.$$

Image of object at $$\infty $$ must lie within distance upto which person can view clearly.
Image distance, $$v = 400\,cm = 4\,m \Rightarrow u = \infty $$
Using lens equation, $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\eqalign{ & \Rightarrow \frac{1}{{ - 4}} - \frac{1}{\infty } = \frac{1}{f} \cr & \Rightarrow f = - 4\,m \cr} $$
Now power of the required lens is,
$$P = \frac{1}{f} = \frac{1}{{ - 4}} = - 0.25D$$
Thus, the person require a concave lens of power $$ - 0.25D.$$

$$\eqalign{ & P = 2\left[ {\frac{{100}}{{10}} + \frac{{100}}{{ - 20}} + 0} \right] \cr & P = 10\,{\text{dioptre}}{\text{.}} \cr} $$

Given, radius of hemispherical glass
$$R = 10\,cm$$
$$\therefore $$ Focal length $$f = \frac{{10}}{2} = - 5\,cm$$
$$u = \left( {10 - 6} \right) = - 4\,cm.$$
By using mirror formula,
$$\eqalign{ & \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \cr & \Rightarrow \frac{1}{v} + \frac{1}{{ - 4}} = \frac{1}{{ - 5}} \cr & \Rightarrow v = 20\,cm. \cr} $$
Apparent height, $${h_a} = {h_r}\frac{{{\mu _1}}}{{{\mu _2}}} = 30 \times \frac{1}{{1.5}} = 20\,cm.$$
below flat surface.

We know, magnification of telescope, we have
$$\eqalign{ & M = \frac{{{f_0}}}{{{f_e}}}, \cr & {\text{Here,}}\,\,\frac{{{f_e}}}{{{f_e} + u}} = \frac{{ - I}}{L} \cr & \Rightarrow \frac{{{f_e}}}{{{f_e} - \left( {{f_o} + {f_e}} \right)}} = \frac{{ - I}}{L} \cr & \Rightarrow \frac{{{f_e}}}{{{f_o}}} = \frac{I}{L} \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,M = \frac{L}{I} \cr} $$

The incident angle is 45°.
Incident angle > critical angle, $$i > {i_c}$$
$$\eqalign{ & \therefore \,\,\sin i > \sin {i_c}\,\,{\text{or }}\sin {45^ \circ } > \sin {i_c}\left( {\sin {i_c} = \frac{1}{n}} \right) \cr & \therefore \,\,\sin {45^ \circ } > \frac{1}{n}\,\,{\text{or }}\frac{1}{{\sqrt 2 }} > \frac{1}{n} \cr & \Rightarrow \,\,n > \sqrt 2 \cr} $$

Initially, the focal length of equiconvex lens is
$$\eqalign{ & \frac{1}{f} = \left( {\mu - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)\,......\left( {\text{i}} \right) \cr & \frac{1}{f} = \left( {\mu - 1} \right)\left( {\frac{1}{R} - \frac{1}{{ - R}}} \right) = \frac{{2\left( {\mu - 1} \right)}}{R} \cr} $$
Case I
When lens is cut along $$XOX',$$  then each half is again equiconvex with
$$\eqalign{ & {R_1} = R,{R_2} = - R \cr & {\text{Thus,}}\,\,\frac{1}{f} = \left( {\mu - 1} \right)\left[ {\frac{1}{R} - \frac{1}{{\left( { - R} \right)}}} \right] \cr & = \left( {\mu - 1} \right)\left[ {\frac{1}{R} + \frac{1}{R}} \right] \cr & = \left( {\mu - 1} \right)\frac{2}{R} \cr & = \frac{1}{{f'}} \cr & \Rightarrow f' = f \cr} $$
Case II
When lens is cut along $$YOY',$$  then each half becomes planoconvex with
$$\eqalign{ & {R_1} = R,{R_2} = \infty \cr & {\text{Thus,}}\,\,\frac{1}{{f''}} = \left( {\mu - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right) \cr & = \left( {\mu - 1} \right)\left( {\frac{1}{R} - \frac{1}{\infty }} \right) \cr & = \frac{{\left( {\mu - 1} \right)}}{R} \cr & = \frac{1}{{2f}} \cr & {\text{Hence,}}\,\,f' = f,f'' = 2f \cr} $$
NOTE
When we cut a convex lens along principal axis both the lens formed are convex lens of same focal length as that of original.
When we cut a convex lens vertical to principal axis, each lens has focal length twice of original.

KEY CONCEPT : The focal length $$(F)$$  of the final mirror is
$$\frac{1}{F} = \frac{2}{{{f_\ell }}} + \frac{1}{{{f_m}}}$$
$$\eqalign{ & {\text{Here, }}\frac{1}{{{f_\ell }}} = \left( {\mu - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right) \cr & = \left( {1.5 - 1} \right)\left[ {\frac{1}{\alpha } - \frac{1}{{ - 30}}} \right] \cr & = \frac{1}{{60}} \cr & \therefore \,\,\frac{1}{F} = 2 \times \frac{1}{{60}} + \frac{1}{{\frac{{30}}{2}}} \cr & = \frac{1}{{10}} \cr & \therefore \,\,F = 10\,cm \cr} $$
The combination acts as a converging mirror. For the object to be of the same size of mirror,
$$u = 2F = 20\,cm$$

$$\eqalign{ & \frac{y}{D} \geqslant 1.22\frac{\lambda }{d} \cr & \Rightarrow \,\,D \leqslant \frac{{yd}}{{\left( {1.22} \right)\lambda }} \cr & = \frac{{{{10}^{ - 3}} \times 3 \times {{10}^{ - 3}}}}{{\left( {1.22} \right) \times 5 \times {{10}^{ - 7}}}} \cr & = \frac{{30}}{{6.1}} \cr & \approx 5\,m \cr & \therefore \,\,{D_{\max }} = 5\,m \cr} $$

Image speed $$ = - \left( {\frac{{{v^2}}}{{{u^2}}}} \right)\frac{{du}}{{dt}}$$
For convex mirror $$\left| {\frac{v}{u}} \right| < 1$$
∴ Image speed is always less than $$v.$$