Q151 / 170 · Manipal (MET) 2019 · Physics
A point object $O$ is placed in front of a glass rod having spherical end of radius of curvature 30 cm . The image would be formed at
- Correct answerA30 cm left
- Binfinity
- C1 cm to the right
- D18 cm to the left
Solution
Focal length of glass rod, $f=\frac{R}{2}$
$=\frac{30}{2}=15 \mathrm{~cm}$
Distance of object, $u=-15 \mathrm{~cm}$ From formula, $\begin{aligned} \frac{\mu_2}{v}-\frac{\mu_1}{u} & =\frac{\mu_2-\mu_1}{R} \\ \mu_2 & =1.5 \quad \text { (for glass) } \\ \mu_1 & =1 \quad \text { (for air) } \\ \frac{1.3}{v} & =-\frac{1}{-15}=\frac{1.5-1}{30} \\ \frac{1.5}{v}+\frac{1}{15} & =\frac{0.5}{30} \\ \frac{1.5}{v} & =\frac{1}{60}-\frac{1}{15} \\ \frac{1.5}{v} & =\frac{-1}{20} \\ v & =-30 \mathrm{~cm} \end{aligned}$
Distance of object, $u=-15 \mathrm{~cm}$ From formula, $\begin{aligned} \frac{\mu_2}{v}-\frac{\mu_1}{u} & =\frac{\mu_2-\mu_1}{R} \\ \mu_2 & =1.5 \quad \text { (for glass) } \\ \mu_1 & =1 \quad \text { (for air) } \\ \frac{1.3}{v} & =-\frac{1}{-15}=\frac{1.5-1}{30} \\ \frac{1.5}{v}+\frac{1}{15} & =\frac{0.5}{30} \\ \frac{1.5}{v} & =\frac{1}{60}-\frac{1}{15} \\ \frac{1.5}{v} & =\frac{-1}{20} \\ v & =-30 \mathrm{~cm} \end{aligned}$
Hence, the correct answer is option: 'A'.