Question

An eqiconvex lens has a power of 5D.if it is made of refractive index 1.5 then radius of curvature of its each surface will be?

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\red{\boxed{\sf Radius\:of\: Curvature\:(R)=20\:cm }}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Power (P) = 5D

• Refractive index $\sf{(\mu)=1.5}$

$\large\underline\pink{\sf To\:Find: }$

• Radius of curvature (R) = ?

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$\large{\boxed{\sf P=\frac{1}{f}}}$

$\large\implies{\sf f=\frac{1}{P}}$

$\large\implies{\sf f=\frac{1}{5}}$

$\large\implies{\sf f=0.2m}$

$\large\implies{\sf f=0.2×100}$

$\large\implies{\sf f=20cm}$

For Equiconvex lens :-

$\large\implies{\sf \frac{1}{f}=(\mu-1)\left(\frac{1}{+R}-\frac{1}{-R}\right) }$

$\large\implies{\sf \frac{1}{f}=(\mu-1)\left(\frac{2}{R}\right) }$

$\large{\boxed{\sf f=\frac{R}{2(\mu-1)}}}$

On putting value :

$\large\implies{\sf 20=\frac{R}{2(1.5-1)}}$

$\large\implies{\sf 20=\frac{R}{2×0.5} }$

$\large\implies{\sf 20=\frac{R}{1} }$

$\large\implies{\sf R=20×1 }$

$\large\implies{\sf R=20\:cm}$

$\red{\boxed{\sf Radius\:of\: Curvature\:(R)=20\:cm }}$

Radius of curvature (R) of eqiconvex lens is 20cm