Question

The focal length of a biconvex lens of refractive
index 1.5 is 0.06 m. Radii of curvature are in the ratio 1:2. Then radii of curvature of two lens surfaces are​

Answer 1

Answer:

As, 1f=(μ−1)(1R1−1R2)

∴10.06=(1.5−1)(1R1+1R2)

(Taking R1 positive and R2 negative)

or 1R1+1R2=10.006×0.5=1003

According to question, R1R2=12orR2=2R1

∴32R1=1003

R1=9200=0.045m

R2=2R1=2×0.045m=0.09m

Explanation:

Answer 2

Answer:

The radii of curvature of two lens surfaces are 0.045m and 0.09m.

Explanation:

Given the focal length of a biconvex lens, f = 0.06 m

Refractive index of the lens, $\mu=1.5$

Ratio of radii of curvatures, $R_{1}: R_{2}=1:2$

$\implies R_{2}=2 R_{1}$

Using the lens maker's formula,

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

For biconvex lens, $R_{2}$ is negative.

And substituting the values,

$\frac{1}{0.06} =(1.5-1)\Big(\frac{1}{R_{1} } +\frac{1}{{2R_{1} }}\Big)$

$\implies \frac{1}{0.06} =(0.5)\Big(\frac{2R_{1}+R_{1}}{{2R_{1} R_{1}}}\Big)$

$\implies \frac{1}{0.06\times 0.5}=\frac{3}{{2R_{1} }}$

$\implies2R_{1}= 0.06\times 0.5\times3$

$\implies 2R_{1}= 0.09m$

$\implies R_{1}= \frac{0.09}{2} =0.045m$

Therefore, the radii of curvature of two lens surfaces are 0.045m and 0.09m.