Question

A double concave lens of refractive index 1.5 with its two spherical surfaces of radii R₁ = 30 cm and R₂ = 45 cm is kept in air. Its focal length is

• A. 15 cm
• B. 37.5 cm
• C. -18 cm
• D. -36 cm

Answer 1

The Correct Option is D

Step 1: Recall the lens maker's formula.

The lens maker's formula for a thin lens is:

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

where:

• \( f \) is the focal length,
• \( n \) is the refractive index of the lens material,
• \( R_1 \) and \( R_2 \) are the radii of curvature of the two surfaces.

 

Step 2: Substitute the given values.

For a double concave lens:

• \( R_1 = -30 \, \text{cm} \) (negative because the surface is concave),
• \( R_2 = 45 \, \text{cm} \) (positive because the surface is concave),
• \( n = 1.5 \).

Substitute into the formula:

 

\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{-30} - \frac{1}{45} \right). \]

Simplify:

\[ \frac{1}{f} = 0.5 \left( -\frac{1}{30} - \frac{1}{45} \right). \]

Find a common denominator for the fractions:

\[ -\frac{1}{30} - \frac{1}{45} = -\frac{3}{90} - \frac{2}{90} = -\frac{5}{90} = -\frac{1}{18}. \]

Substitute back:

\[ \frac{1}{f} = 0.5 \cdot \left( -\frac{1}{18} \right) = -\frac{1}{36}. \]

Invert to find \( f \):

\[ f = -36 \, \text{cm}. \]

Final Answer: The focal length is \( \mathbf{-36 \, \text{cm}} \), which corresponds to option \( \mathbf{(4)} \).