Calculate the radius of curvature of an equi-concave lens of refractive index 1.5, when it is kept in a medium of refractive index 1.4, to have a power of -5\; D ?

 

 

 

 
 
 
 
 

Answers (1)

The radius can be taken 'R'    [ or, Let the radius be 'R']

then, R_{1}=-R and R_{2}=R

According to question

Power of Lens, = -5D

Refractive index

\\n_2=1.5\\n_1=1.4

we know the lens makers formula  -

P=\frac{1}{f}=(\frac{n_2}{n_1}-1)\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )

On putting the values we have - 

\\-5m^{-1}=(\frac{1.5}{1.4}-1)(\frac{-1}{R}-\frac{1}{R})\\\\\Rightarrow 5=\frac{0.2}{1.4R}\\\\\Rightarrow R=\frac{0.2}{7}=0.0286 m=2.86 cm

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