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

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Q.
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 : [2025]

1 $\frac{1}{3} cm and \frac{1}{3} cm$

2 $\frac{1}{5} cm and \frac{1}{7} cm$

3 $\frac{1}{3} cm and \frac{1}{7} cm$

4 $\frac{1}{6} cm and \frac{1}{9} cm$

Ans.
(2)

$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

In 1st case:

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

$= (μ - 1) (\frac{2}{R}) = (μ - 1) 12$

in 2nd case:

$\frac{1}{f'} = (μ - 1) (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

If $f' = f \Rightarrow 12 = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

Hence, option (2) is correct.

Solution

1 $\frac{1}{3} cm and \frac{1}{3} cm$

2 $\frac{1}{5} cm and \frac{1}{7} cm$

3 $\frac{1}{3} cm and \frac{1}{7} cm$

4 $\frac{1}{6} cm and \frac{1}{9} cm$

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Solution

Q. 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 (R1≠R2), without any change in lens power then possible combination of R1 and R2 is : [2025]

1 13 cm and 13 cm

2 15 cm and 17 cm

3 13 cm and 17 cm

4 16 cm and 19 cm

Ans. (2) 1f=(μ–1)(1R1–1R2) In 1st case: 1f=(μ–1)(1R–1–R) =(μ–1)(2R)=(μ–1)12 in 2nd case: 1f'=(μ–1)(1R1+1R2) Iff'=f ⇒ 12=1R1+1R2 Hence, option (2) is correct.

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1 $\frac{1}{3} cm and \frac{1}{3} cm$

2 $\frac{1}{5} cm and \frac{1}{7} cm$

3 $\frac{1}{3} cm and \frac{1}{7} cm$

4 $\frac{1}{6} cm and \frac{1}{9} cm$