Question
A thin lens, made of material of refractive index μ has a focal length f. If the lens is placed in a transparent medium of refractive index ‘n’ (n < μ), obtain an expression for the change in focal length of the lens. Use the result to show that the focal length of a lens of the glass = becomes times its focal length in air, when it is placed in water (μ = μw).
Solution
Here, we have a thin lens of refractive index '' and focal length 'f'. This lens is placed in a transparent medium of refractive index 'n'.
Let, the radii if curvature of the lens be R1 and R2.
Therefore, using lens makers formula , focal length is given by,
...(i)
When the lens is put in a transparent medium, the refractive index of the material of the lens with respect to the medium
Thus focal length f' of the lens in the medium is given by,
...(ii)
Now, from Eqns. (i) and (ii),
i.e., ...(iii)
Thus, change in focal length is,
...(iv)
From equation (iii), we have
Putting,
where,
is the refractive index of glass and,
is te refractive index of water.
The ray diagram is as shown below:
Let, the radii if curvature of the lens be R1 and R2.
Therefore, using lens makers formula , focal length is given by,
...(i)
When the lens is put in a transparent medium, the refractive index of the material of the lens with respect to the medium
Thus focal length f' of the lens in the medium is given by,
...(ii)
Now, from Eqns. (i) and (ii),
i.e., ...(iii)
Thus, change in focal length is,
...(iv)
From equation (iii), we have
Putting,
where,
is the refractive index of glass and,
is te refractive index of water.
The ray diagram is as shown below:
