A convex lens of refractive index $\frac{3}{2}$ has a power of $2.5 \mathrm{D}$ in air. If it is placed in a liquid of refractive index 2 , then the new power of the lens is

• Correct answer
  A

  $-1.25 \mathrm{D}$
• B

  $-1.5 \mathrm{D}$
• C

  $1.25 \mathrm{D}$
• D

  $1.5 \mathrm{D}$

Solution

Focal length of a convex lens having power $2.5 \mathrm{D},=\frac{1}{2.5} \mathrm{~m}$ Also focal length of a lens in a medium of refractive index $\mu$ is given by $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Dividing the two, $2.5 \frac{0.5}{-0.25}$ $\Rightarrow \frac{1}{f^{\prime}}=\frac{-5}{25 \times 0.25}=-1.25 \mathrm{D}$

Hence, the correct answer is option: 'A'.