10. A lens is required to have a power of -2.5 diopters in air. The convex front surface has a radius of curvature of 3.0 cm. What is the radius of curvature of the rear surface?

A. 0.12 cm
B. 1.2 cm
C. 12 cm
D. -0.012 cm
E. 120 cm



Answer :

To solve the given problem, let's break it down step-by-step using the Lensmaker's equation:

The power ([tex]\(P\)[/tex]) of the lens is given as [tex]\(-2.5\)[/tex] diopters. The radius of curvature of the convex front surface ([tex]\(R_1\)[/tex]) is [tex]\(3.0\)[/tex] cm. The refractive index ([tex]\(n\)[/tex]) for the glass lens in air is typically [tex]\(1.5\)[/tex].

The Lensmaker's equation relates the radii of curvature of the two surfaces of the lens, the refractive index of the lens material, and the focal length of the lens. The equation is:

[tex]\[ \frac{1}{f} = (n - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \][/tex]

Where:
- [tex]\( f \)[/tex] is the focal length of the lens.
- [tex]\( R_1 \)[/tex] is the radius of curvature of the front surface.
- [tex]\( R_2 \)[/tex] is the radius of curvature of the rear surface.
- [tex]\( n \)[/tex] is the refractive index of the lens material.

Since the power [tex]\( P \)[/tex] of the lens is given by:

[tex]\[ P = \frac{1}{f} \][/tex]

We can substitute [tex]\( P \)[/tex] into the Lensmaker's equation to find [tex]\( R_2 \)[/tex]:

[tex]\[ P = (n - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \][/tex]

Rearranging this equation to solve for [tex]\( R_2 \)[/tex]:

1. First, calculate the focal length ([tex]\( f \)[/tex]) in centimeters.

[tex]\[ f = \frac{1}{P} = \frac{1}{-2.5} \text{ diopters} \][/tex]

Since 1 diopter equals 100 cm in terms of focal length:

[tex]\[ f = \frac{100}{-2.5} = -40 \text{ cm} \][/tex]

2. Substitute the given values into the modified Lensmaker's equation and solve for [tex]\( R_2 \)[/tex]:

[tex]\[ -2.5 = (1.5 - 1) \left(\frac{1}{3.0} - \frac{1}{R_2}\right) \][/tex]

3. Simplify the equation:

[tex]\[ -2.5 = 0.5 \left(\frac{1}{3.0} - \frac{1}{R_2}\right) \][/tex]

4. Multiply both sides by 2 to clear the fraction:

[tex]\[ -5 = \left(\frac{1}{3.0} - \frac{1}{R_2}\right) \][/tex]

5. Rearrange to isolate [tex]\( \frac{1}{R_2} \)[/tex]:

[tex]\[ \frac{1}{R_2} = \frac{1}{3.0} + 5 \][/tex]

6. Calculate each term:

[tex]\[ \frac{1}{3.0} = 0.3333\][/tex]
[tex]\[ 0.3333 + 5 = 5.3333\][/tex]

7. Take the reciprocal to find [tex]\( R_2 \)[/tex]:

[tex]\[ R_2 = \frac{1}{5.3333} \approx 0.1875 \text{ cm} \][/tex]

Thus, the radius of curvature of the rear surface is approximately [tex]\( 0.1875 \)[/tex] cm.

The correct answer is not explicitly listed among the choices, but [tex]\(0.1875\)[/tex] cm is the accurate answer calculated based on the given conditions.