Physics Question on Ray optics and optical instruments

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

Answer 1

Solution

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.