Q145 / 160 · AP EAMCET 2021 (24 Aug Shift 1) · Physics
The radius of curvature of the face of planoconvex lens is $12 \mathrm{~cm}$ and its refractive index is 1.5. Then, the focal length of the lens is
- A$26 \mathrm{~cm}$
- B$22 \mathrm{~cm}$
- Correct answerC$24 \mathrm{~cm}$
- D$20 \mathrm{~cm}$
Solution
Given, refractive index of lens, $\mu=1.5$
Radius of curved face, $R_1=12 \mathrm{~cm}$
Radius of plane face, $R_2=\infty$

Now, using the expression of focal length, $$ \frac{1}{f}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] $$ Substituting the values, we get $$ \begin{aligned} \frac{1}{f} & =(1.5-1)\left[\frac{1}{12}-\frac{1}{\infty}\right] \\ \frac{1}{f} & =0.5 \times \frac{1}{12} \\ \Rightarrow \quad f & =\frac{12}{0.5} \mathrm{~cm}=24 \mathrm{~cm} \end{aligned} $$ Hence, the focal length of given plano-convex lens is $24 \mathrm{~cm}$.

Now, using the expression of focal length, $$ \frac{1}{f}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] $$ Substituting the values, we get $$ \begin{aligned} \frac{1}{f} & =(1.5-1)\left[\frac{1}{12}-\frac{1}{\infty}\right] \\ \frac{1}{f} & =0.5 \times \frac{1}{12} \\ \Rightarrow \quad f & =\frac{12}{0.5} \mathrm{~cm}=24 \mathrm{~cm} \end{aligned} $$ Hence, the focal length of given plano-convex lens is $24 \mathrm{~cm}$.
Hence, the correct answer is option: 'C'.