Q144 / 160 · TS EAMCET 2005 · Physics
The focal length of an equi-convex lens is greater than the radius of curvature of any of the surfaces. Then the refractive index of the material of the lens is
- Correct answerAgreater than zero but less than 1.5
- Bgreater than 1.5 but less than 2.0
- Cgreater than 2.0 but less than 2.5
- Dgreater than 2.5 but less than 2.0
Solution
Focal length of lens,
$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
For equi-convex lens,
$\begin{aligned}
R_1 & =+R, R_2=-R \\
\frac{1}{f} & =(\mu-1)\left(\frac{1}{R}-\frac{1}{-R}\right) \\
\frac{1}{f} & =(\mu-1)\left(\frac{2}{R}\right) \\
f & =\frac{R}{2(\mu-1)} \\
f < R, \text { so, } 2(\mu-1) & < 1 \\
(\mu-1) & < \frac{1}{2} \\
\text { i.e. } \quad(\mu-1) & < 0.5 \\
\mu & < 1.5
\end{aligned}$
Focal length of convex lens is positive. So, $\mu$ cannot be negative, hence $\mu$ should be greater than zero but less than 1.5.
Hence, the correct answer is option: 'A'.