To determine the relative decrease in the focal length of a lens when the optical power changes, we need to understand the relationship between optical power and focal length.
The optical power \( P \) of a lens is given by the formula:
\(P = \frac{1}{f}\)
where:
Initially, the optical power of the lens is \( P_1 = 2.5 \, \text{D} \). Therefore, the focal length \( f_1 \) is:
\(f_1 = \frac{1}{P_1} = \frac{1}{2.5} = 0.4 \, \text{m}\)
The optical power is increased by \( 0.1 \, \text{D} \) to become \( P_2 = 2.6 \, \text{D} \). The new focal length \( f_2 \) is:
\(f_2 = \frac{1}{P_2} = \frac{1}{2.6} \approx 0.3846 \, \text{m}\)
The decrease in focal length is:
\(\Delta f = f_1 - f_2 = 0.4 - 0.3846 = 0.0154 \, \text{m}\)
The relative decrease in focal length is calculated by the formula:
\(\text{Relative decrease} = \frac{\Delta f}{f_1} = \frac{0.0154}{0.4} = 0.0385\)
Rounding to two decimal places, the relative decrease is approximately 0.04.
Therefore, the correct answer is 0.04.
The relationship between optical power \( P \) and focal length \( f \) is given by:
\[ P = \frac{1}{f}. \] Thus, the initial focal length is: \[ f_1 = \frac{1}{P_1} = \frac{1}{2.5} = 0.4 \, \text{m}. \]
After the optical power increases by 0.1 D, the new optical power becomes:
\[ P_2 = 2.5 + 0.1 = 2.6 \, \text{D}. \] The new focal length is: \[ f_2 = \frac{1}{P_2} = \frac{1}{2.6} \approx 0.3846 \, \text{m}. \]
The relative decrease in focal length is: \[ \text{Relative decrease} = \frac{f_1 - f_2}{f_1} = \frac{0.4 - 0.3846}{0.4} = \frac{0.0154}{0.4} \approx 0.04. \]
The correct answer is option (1) \( \boxed{0.04} \).
Light from a point source in air falls on a spherical glass surface (refractive index, \( \mu = 1.5 \) and radius of curvature \( R = 50 \) cm). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is 1cm.