Question

A convex lens of radii of curvature 20 cm and 30 cm respectively. It is silvered at the surface which has smaller radius of curvature. Then if will behave as ():

• A. Concave mirror with equivalent focal length $\frac { 30 } { 11 }$cm
• B. Concave mirror with equivalent focal length$\frac { 60 } { 11 }$cm
• C. Concave mirror with equivalent focal length$\frac { 30 } { 11 }$cm.
• D. Concave mirror with equivalent focal length$\frac { 60 } { 11 }$cm.

Answer 1

Answer: (B)

Concave mirror with equivalent focal length$\frac { 60 } { 11 }$cm

Answer 2

: Focal length for lens

$\frac { 1 } { \mathrm { f } _ { \mathrm { L } } } = \left( \mu _ { \mathrm { g } } - 1 \right) \left( \frac { 1 } { \mathrm { R } _ { 1 } } - \frac { 1 } { \mathrm { R } _ { 2 } } \right)$

Here, $\mathrm { u } _ { \mathrm { g } } = 1.5 , \mathrm { R } _ { 1 } = 20 \mathrm {~cm} , \mathrm { R } _ { 2 } = - 30 \mathrm {~cm}$

$= ( 1.5 - 1 ) \left( \frac { 1 } { 20 } - \frac { 1 } { - 30 } \right) = 0.5 \times \left( \frac { 5 } { 60 } \right)$

or $\mathrm { f } _ { \mathrm { L } } = \frac { 120 } { 5 } = 24 \mathrm {~cm}$

Focal length for mirror

$f _ { m } = \frac { R } { 2 } = \frac { - 20 } { 2 } = - 10 \mathrm {~cm}$

$\therefore$ Equivalent focal length

$\Rightarrow \mathrm { f } _ { \mathrm { eq } } = \frac { - 60 } { 11 } \mathrm {~cm}$

Hence, this system begaves like a concave mirror of focal length $\frac { - 60 } { 11 } \mathrm {~cm}$