(a) With the help of a suitable ray diagram, derive the mirror formula for a concave mirror. (b) The near point of a hypermetropic person is $50\ \mathrm{cm}$ from the eye. What is the power of the lens required to enable the person to read clearly a book held at $25\ \mathrm{cm}$ from the eye?

(a) $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$. (b) Required power $=+2\ \mathrm{D}$ (a converging lens).

CBSE Class 12 · 2009 · Geometrical OpticsLong Answer

(a) With the help of a suitable ray diagram, derive the mirror formula for a concave mirror.
(b) The near point of a hypermetropic person is

50 cm

from the eye. What is the power of the lens required to enable the person to read clearly a book held at

25 cm

from the eye?

Answer: (a) $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ .
(b) Required power $= + 2 D$ (a converging lens).

(a) Using similar triangles from the ray diagram of a concave mirror (object

A B

, image

A^{'} B^{'}

, pole

P

, focus

F

\frac{A ^{'} B ^{'}}{A B} = \frac{P B ^{'}}{P B}

and

\frac{A ^{'} B ^{'}}{A B} = \frac{B ^{'} F}{P F}

. Equating and applying the sign convention (

P B = - u

P B^{'} = - v

P F = - f

) gives

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

(b) The lens must image the book (at

25 cm

) at the person's near point (

50 cm

u = - 25 cm

v = - 50 cm

\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{- 50} - \frac{1}{- 25} = \frac{1}{50} \Rightarrow f = + 50 cm, P = \frac{100}{50} = + 2 D

CBSE Class 12 Board · 2009 · Subjective