Question

One plano-convex and one plano-concave lens of same radius of curvature ‘R’ but of different materials are joined side by side as shown in the figure. If the refractive index of the material of 1 is μ₁ and that of 2 is μ₂, then the focal length of the combination is (A) R/2(μ₁- μ₂)
(B) 2R/(μ₁- μ₂)
(C) R/(μ₁- μ₂)
(D) R/{2-(μ₁ - μ₂)}

Answer 1

Explanation:

One plano-convex and one plano-concave lens of same radius of curvature ‘R’ but of different materials are joined side by side as shown in the figure. If the refractive index of the material of 1 is μ₁ and that of 2 is μ₂, then the focal length of the combination is (A) R/2(μ₁- μ₂)

(B) 2R/(μ₁- μ₂)

(C) R/(μ₁- μ₂)

(D) R/{2-(μ₁ - μ₂)}

Answer 2

Answer:

The focal length of the combination of of a plano-convex and plano-concave lens is $f=\frac{R}{\mu_{1}-\mu_{2}}$

Explanation:

Given, the radius of curvature of a plano-convex and plano-concave lens is  R.

refractive index of plano-convex lens is $\mu_{1}$

and of plano-concave lens is $\mu_{2}$.

The focal length of plano-convex is

                       $\frac{1}{f_{1} }=(\mu_{1}-1)\big(\frac{1}{\infty }-\frac{1}{(-R) }\big)$

                           $=\frac{\mu_{1}-1}{R }$

The focal length of plano-concave is

                      $\frac{1}{f_{2} }=(\mu_{2}-1)\big(\frac{1}{(-R)}-0 }\big)$

                          $=-\frac{\mu_{2}-1}{R}$

Then the combined focal lengths is

                       $\frac{1}{f}= \frac{1}{f_{1} }+\frac{1}{f_{2} }$

                          $=\frac{\mu_{1}-1}{R}-\frac{\mu_{2}-1}{R}$

                          $=\frac{\mu_{1}-\mu_{2}}{R}$

Therefore, the focal length of the combination of lens. $f=\frac{R}{\mu_{1}-\mu_{2}}$

So, the correct answer is option 3.