A biconvex lens has radius of curvature of both the surfaces same as 6 1 cm. If this lens is replaced by another convex lens having different radii of curvature on both sides without change in lens

Question

A biconvex lens has radius of curvature of both the surfaces same as 6 1 ​ cm. If this lens is replaced by another convex lens having different radii of curvature on both sides without change in lens power then possible combination of R 1 ​ and R 2 ​ is ( R 1 ​  = R 2 ​ )

TestHub · Accepted Answer

Using the Lens Maker's formula, the power of a lens is given by:

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

For the initial biconvex lens, $R_{1} = \frac{1}{6}$ cm and $R_{2} = - \frac{1}{6}$ cm. Substituting these values:

$P = (μ - 1) (6 - (- 6)) = 12 (μ - 1)$

For the new biconvex lens, the power remains the same. Let its radii of curvature be $R_{1}^{'}$ and $R_{2}^{'}$ (where $R_{1}^{'}$ is positive and $R_{2}^{'}$ is negative). Let their magnitudes be $r_{1}$ and $r_{2}$ .

$P^{'} = (μ - 1) (\frac{1}{r _{1}} + \frac{1}{r _{2}})$

Equating the powers $P = P^{'}$ :

$\frac{1}{r _{1}} + \frac{1}{r _{2}} = 12$

Checking the given options for the sum of the reciprocals of the radii:

For $1/3$ cm and $1/5$ cm: $3 + 5 = 8 \neq = 12$

For $1/5$ cm and $1/7$ cm: $5 + 7 = 12$

For $1/3$ cm and $1/7$ cm: $3 + 7 = 10 \neq = 12$

For $1/6$ cm and $1/9$ cm: $6 + 9 = 15 \neq = 12$

The combination of $1/5$ cm and $1/7$ cm satisfies the condition.

Answer: $1/5$ cm and $1/7$ cm

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