Question

The refractive index of a lens material is $\mu$ and focal length f. Due to some chemical changes in the material, its refractive index has increased by 2%. The percentage decrease, in focal length for $\mu =1.5$ will be:

• A. $4$%
• B. $2$%
• C. $5.6$%
• D. $8$%

Answer 1

The correct option is C $5.6$%

From lens makers formula,$\frac{1}{f}=[\mu -1][\frac{1}{R_1}-\frac{1}{R_2}]=\frac{1}{2}[\frac{1}{R_1}-\frac{1}{R_2}]$

$\Rightarrow \frac{1}{f_1}=[1.02\mu -1][\frac{1}{R_1}-\frac{1}{R_2}]=\frac{53}{100}[\frac{1}{R_1}-\frac{1}{R_2}]$

$\Rightarrow \frac{|f-f_1|}{f}=\frac{2-\frac{100}{53}}{2}\times 100$

$\Rightarrow \frac{106-100}{53\times 2}\times 100=\frac{6\times 100}{106}$

$=5.6$%