Question

A convex lens of refractive index 1.5 has radii of curvature +20 cm and $-$20 cm. The focal length of the lens is:

• A. 10 cm
• B. 20 cm
• C. 40 cm
• D. 80 cm

Hint:

For an equiconvex lens (where $R_1 = +R$ and $R_2 = -R$) made of glass with refractive index $\mu = 1.5$:
Substituting into the formula gives:
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{R} + \frac{1}{R} \right) = 0.5 \left( \frac{2}{R} \right) = \frac{1}{R} \implies f = R \]
So, the focal length is exactly equal to the radius of curvature!
Recognizing this pattern helps solve such problems instantly.

Answer 1

The Correct Option is B

Step 1: Lensmaker's equation.
$1/f=(n-1)(1/R_1-1/R_2)=0.5\times2/20=0.05$
\[ \boxed{f = 20 \text{ cm}} \]