## Given: A thin biconvex lens of refractive index 3/2 and radius of curvature 50 cm is placed on a reflecting convex surface of radius of curvature 100 cm. A point object is placed on the principal axis of the system such that its final image coincides with itself. Now, few drops of a transparent liquid is placed between the mirror and lens such that final image of the object is at infinity.

To find refractive index of the liquid used. Also, find the position of the object

Solution:

The system behaves as mirror

Power of lens,

$P_{l}=f_{l}1 =(23 −1)(501 −−501 )=501 $

Therefore power of equivalent mirror

$P_{EM}=2P_{l}+P_{m}=502 −501 =501 $

i.e., Focal length of equivalent mirror, $f_{cm}=−50$

Since image and the object coincides, hence position of the object is at the center of curvature of the equivalent mirror, i.e., at $2f_{cm}=100cm$ from lens

When the liquid of refractive index be is placed between the lens and convex mirror, if forms a liquid lens of power,

$P_{u}=(−501 −1001 )=1003(1−μ) $

Now, power of equivalent mirror

$2P_{l}+2P_{u}+P_{m}⟹=502 +1002×3(1−μ) +−501 =503(1−μ)+1 ...........(i)$

Since final image is at infinity, hence the object should be at its focal point

i.e., $f_{EM}=100⟹f_{EM}1 =1001 ...........(ii)$

From eqn(i) and (ii), we get

$503(1−μ)+1 =1001 ⟹6(1−μ)+2=1⟹6−6μ=−1⟹μ=67 $

is the required refractive index of the liquid used.