The two surfaces of a concave lens, made of glass of refractive index 1.5 have the same radii of curvature…

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
  1. becomes a convergent lens of focal length $3.5 R$
  2. becomes a convergent lens of focal length $3.0 R$
  3. changes as a divergent lens of focal length $3.5 R$
  4. 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

Asked in: AP EAMCET 2013

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