A thin equiconvex lens of refractive index `3//2` is placed on a horizontal plane mirror as shown in figure. The space between the lens and the mirror is filled with a liquid of refractive index `4//3` . It is found that when a point object is placed 15 cm above the lens on its priincipal axis, the object coincides with its own image.

Q. If another liquid is filled instead of water, the object and the image coincide at a distance 25 cm from the lens.
Calculate the refractive index of the liquid.

Answer

A

Explanation

The lens makes formula is
`(1)/(f)=(mu-1)((1)/(R_1)-(1)/(R_2))`
When the space between the lens and the mirror is filled with water, a system of two lenses is formed.
(i) a glass lens
(ii) a plano concave water lens
For glass lens Here `R_1=+R` and `R_2=-R` `(1)/(f_e)=(1.5-1)((1)/(R)-(1)/(-R))=(1)/(R)`
For water lens
`(1)/(f_w)=(1.33-1)((1)/(-R)-(1)/(-infty))=(-0.33)/(R)`
The focal length of the combination of two lenses will be
`(1)/(f)=(1)/(f_1)+(1)/(f_2)=(1)/(R)-(0.33)/(R)=(0.67)/(R) `...(ii)
A convex lens placed on a plane mirror behave like a concave mirror. The image is formed at the object itself if the object is placed at centre of curvature of concave mirror. After refraction through lens, the rays fall on the plane mirror normally and retrace their path to form image at the oobject itself.
`:'` Focal length of system`(f)=15`cm` ...(ii)
From (i) and (ii)
`(1)/(15)=(0.67)/(R)impliesR=10.05`cm
The same situation is repeated with two differences
(a) The object and image distance are no 25 cm and
(b) In place of water there is a new liquid of refraction index `mu`
Again `(1)/(f_e)=(1)/(R)` and `(1)/(f^(`))=(-(mu-1))/(R)` where `f^` ` is the focal length of new liquid lens. `:'` New combined lens,
`(1)/(F)=(1)/(f_l)+(1)/(f^`)=(1)/(R)-(+(mu-1))/(R)=(1-mu+1)/(R)=(2-mu)/(R)` ...(i)
For new combined lens,
`:'``(1)/(F)=(1)/(25)`
From (i) and (ii)
`(2-mu)/(10.02)=(1)/(25)` `:'``mu=1.6`