JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Geometrical Optics

A bi-convex lens has radius of curvature of both the surfaces same as

1/6

cm. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides (

R_{1} \neq = R_{2}

), without any change in lens power then possible combination of

R_{1}

and

R_{2}

is :

Answer: (B)

Solution

The power of a thin lens is

\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

. For the equi-biconvex lens the first surface is convex with

R_{1} = + \frac{1}{6}

cm and the second is concave towards the object with

R_{2} = - \frac{1}{6}

cm, so

\frac{1}{R _{1}} - \frac{1}{R _{2}} = 6 + 6 = 12 c m^{- 1}

. Keeping the same power with unequal radii needs

\frac{1}{R _{1}} - \frac{1}{R _{2}} = 12

again, so for a biconvex shape the two curvatures must add to

12 c m^{- 1}

. Taking magnitudes

\frac{1}{5}

cm and

\frac{1}{7}

cm gives

5 + 7 = 12

, so the radii

\frac{1}{5}

cm and

\frac{1}{7}

cm reproduce the required power.

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option