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 

• A. - 2.66 cm
• B. - 26.6 cm
• C. 115.44 cm
• D. +36.6 cm

Answer 1

The Correct Option is B

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:

\( \frac{1}{f} = (1.6 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right)\)

\(\frac{1}{f} = 0.6 \times \frac{2}{R}\)

\(\frac{1}{f} = \frac{1.2}{R}\)

Now, we can find the value of \( R \) by rearranging the equation:

\( R = \frac{1.2}{1/f}\)

Substituting the value of \( f \) from the given power:

\( R = 1.2 \times (-4.5 \, \text{D}^{-1})\)

\( R = -5.4 \, \text{m}^{-1}\)

Converting \( R \) to centimeters, we have:

\( R = -5.4 \, \text{m}^{-1} \times 100 \, \text{cm/m}\)

\( R = -540 \, \text{cm}^{-1}\)

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.