Question:

The power of equi - concave lens is - 4.5 D and is made of a material R.I. 1.6, the radii of the curvature of the lens is 

Updated On: Aug 25, 2024
  • - 2.66 cm
  • - 26.6 cm
  • 115.44 cm
  • +36.6 cm
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The Correct Option is B

Solution and Explanation

1/f = (n-1) (1/R1 - 1/R2
Given: 
Power of the lens (P) = -4.5 D 
Material refractive index (n) = 1.6 
Since the lens is equi-concave, the radii of curvature of both surfaces will have the same magnitude but opposite signs (R1 = - R2
Using the lens maker's formula and substituting the given values, we can solve for the magnitude of the radii of curvature: 
1/f = (1.6 - 1)(1/R - 1/-R) 
​1/f = 0.6 x 2/R 
1/f = 1.2/R 
Now, we can find the value of R by rearranging the equation: 
R = 1.2/ 1/f 
R=1.2⋅f. 
Substituting the value of f from the given power, we get: 
R = 1.2 x (- 4.5 D-1
R = - 5.4 m-1 
Converting R to centimeters, we have: 
R = - 5.4 m-1 . 100 cm/m 
R = 540 cm-1 
The radii of curvature of the lens is approximately -540 cm or -5.4 m. Among the given options, the closest value to -540 cm is (B) -26.6 cm.

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