A convex lens of refractive index μ has two surfaces of equal radius of curvature R . The focal length of convex lens is

Correct answer: B. (R)/(2(μ - 1)) Using the lens maker's formula: (1)/(f) = (μ - 1) ( (1)/(R_1) - (1)/(R_2) ) For a biconvex lens with equal radii of curvature, R_1 = +R and R_2 = -R . Substituting these values into the formula: (1)/(f) = (μ -1) ( (1)/(R) - (1)/(-R) ) (1)/(f) = (μ - 1) ( (2)/(R) ) f = (R)/(2(μ - 1)) Answer: (R)/(2(μ - 1))

MHT CET 202622 AprilEvening Shift

Exam: MHT CET
Year: 2026
Date: 22 April
Shift: Evening Shift
Subject: Physics
Chapter: Ray Optics

MHT CET 2026 · 22 April · Evening Shift · Physics · Ray Optics

A convex lens of refractive index μ has two surfaces of equal...

A convex lens of refractive index

\mu

has two surfaces of equal radius of curvature

R

. The focal length of convex lens is

Options

A $\dfrac{2(\mu - 1)}{R}$
B $\dfrac{R}{2(\mu - 1)}$
C $\dfrac{(\mu - 1)}{2R}$
D $\dfrac{2R}{(\mu - 1)}$

Correct answer

B. $\dfrac{R}{2(\mu - 1)}$

Step-by-step solution

Using the lens maker's formula:

\dfrac{1}{f} = (\mu - 1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)<br><br>For a biconvex lens with equal radii of curvature,

R_1 = +R

and

R_2 = -R

.<br><br>Substituting these values into the formula:<br><br>\dfrac{1}{f} = (\mu -1)

\left( \dfrac{1}{R} - \dfrac{1}{-R} \right)

<br><br>

\dfrac{1}{f} = (\mu - 1)

\left( \dfrac{2}{R} \right)<br><br>

f =

\dfrac{R}{2(\mu - 1)}

<br><br>Answer:

\dfrac{R}{2(\mu - 1)}

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