Question: A glass slab of thickness 2 cm and refractive index 2 is placed in contact with a biconvex lens of f...

Question

A glass slab of thickness 2 cm and refractive index 2 is placed in contact with a biconvex lens of focal length 20 cm. The refractive index of the material of the lens is 1.5. The far side of the lens is silvered. Where should an object be placed in front of the slab so that it images on to itself ?

Answer 1

Solution

=

$\frac { 1 } { 20 }$ = (1.5 – 1)

R = 20 cm

=

= $\frac { 2 } { 20 }$ + $\frac { 2 } { R }$

= $\frac { 1 } { 10 } + \frac { 2 } { 20 }$

= $\frac { 4 } { 20 }$

f_eq = $\frac { 20 } { 4 }$ = 5

Shift due to slab = t – = $2 - \frac { 2 } { 2 }$ = 1 cm

Ray coming from I incident normally at concave mirror

\ PI = 2 f_eq = 10 cm

OI = 1 cm

\ PO = 10 + 1 = 11 cm