The two surfaces of a concave lens, made of glass of refractive index 1.5 have the same radii of curvature $R$. It is now immersed in a medium of refractive index 1.75 , then the lens

• Correct answer
  A

  becomes a convergent lens of focal length $3.5 R$
• B

  becomes a convergent lens of focal length $3.0 R$
• C

  changes as a divergent lens of focal length $3.5 R$
• D

  changes as a divergent lens of focal length $3.0 R$

Solution

From lens maker's formula $$ \frac{1}{f}=\left({ }_g^m \mu-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) $$ Now, $\quad{ }_g^m \mu=\frac{g \mu}{{ }_m \mu}=\frac{1.5}{1.75}$ For concave lens as shown in the figure, in this case $$ R_1=-R \text { and } R_2=+R $$
$$ \begin{gathered} \frac{1}{f}=\left(\frac{1.5}{1.75}-1\right)\left(-\frac{1}{R}-\frac{1}{R}\right)=+\frac{0.25 \times 2}{1.75 R} \\ \Rightarrow \quad f=+3.5 R \end{gathered} $$ The positive sign shows that the lens behaves as a convergent lens.

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