Optics Ques 59
- A thin convex lens made from crown glass $(\mu=3 / 2)$ has focal length $f$. When it is measured in two different liquids having refractive indices $4 / 3$ and $5 / 3$, it has the focal lengths $f_1$ and $f_2$, respectively. The correct relation between the focal length is
(2014 Main)
(a) $f _1=f _2<f$
(b) $f_1 > f$ and $f_2$ becomes negative
(c) $f_2 > f$ and $f_1$ becomes negative
(d) $f _1$ and $f _2$ both become negative
Show Answer
Answer:
Correct Answer: 59.(b)
Solution:
- It is based on lens maker’s formula and its magnification.
$$ \text { i.e. } \quad \frac{1}{f}=(\mu-1) \frac{1}{R _1}-\frac{1}{R _2} $$
According to lens maker’s formula, when the lens is in the air
$$ \begin{alignedat} & \frac{1}{f}=\frac{1}{2}-\frac{1}{R _1}-\frac{1}{R _2} \\ & \frac{1}{f}=\frac{1}{2 x} \Rightarrow f=2 x \end{aligned} $$
$$ \text { Here, } \quad \frac{1}{x}=\frac{1}{R _1}-\frac{1}{R _2} $$
In case of liquid, where refractive index is $\frac{4}{3}$ and $\frac{5}{3}$, we get
Focal length of the first liquid
$$ \frac{1}{f _1}=\frac{\mu _s}{\mu _{l _1}}-1 \frac{1}{R _1}-\frac{1}{R _2} \Rightarrow \frac{1}{f _1}=\frac{\frac{3}{2}}{\frac{4}{3}}-1 \frac{1}{x} $$
$\Rightarrow f _1$ is positive.
$$ \begin{alignedat} & \frac{1}{f _1} & =\frac{1}{8 x}=\frac{1}{4(2 x)}=\frac{1}{4 f} \\ ⇒ & f _1 & = 4 f \end{aligned} $$
Focal length in second medium
$$ \frac{1}{f _2}=\frac{\mu _s}{\mu _{l _1}}-1 \quad \frac{1}{R _1}-\frac{1}{R _2} $$
$\Rightarrow \quad \frac{1}{f _2}=\frac{\frac{3}{2}}{\frac{5}{3}}-\frac{1}{x}$
$\Rightarrow f _2$ is negative.
BETA
