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  • The focal length of an equiconcave lens is  times of radius of curvature of its surfaces. Find the refractive index of the material of the lens. Under w

The focal length of an equiconcave lens is \frac{3}{4} times of radius of curvature of its surfaces. Find the refractive index of the material of the lens. Under what condition will this lens behave as a converging lens ?

 

 
 
 
 
 

Answers (1)

\frac{1}{f}=(n-1)[\frac{1}{R_1}-\frac{1}{R_2}]

\\R_1=-R\\R_2=R

\\\frac{-4}{3R}=(n-1)[\frac{-2}{R}]\\implies\\n-1=\frac{2}{3}\\\\n=\frac{5}{3}

If the lens is immersed in a medium of refractive index greater than 5/3 , it will behave like a converging lens

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Safeer PP

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