Optics 7 Question 41
42. A thin equiconvex lens of glass of refractive index $\mu=3 / 2$ and of focal length $0.3 m$ in air is sealed into an opening at one end of a tank filled with water $\mu=4 / 3$. On the opposite side of the lens, a mirror is placed inside the tank on the tank wall perpendicular to the lens axis, as shown in figure. The separation between the lens and the mirror is $0.8 m$. A small object is placed outside the tank in front of lens. Find the position (relative to the lens) of the image of the object formed by the system
(1997C, 5M)
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Answer:
Correct Answer: 42. $0.9 m$ from the lens (rightwards) or $0.1 m$ behind the mirror
Solution:
- From lens maker’s formula,
$$ \frac{1}{f}=(\mu-1) \frac{1}{R _1}-\frac{1}{R _2} $$
we have $\quad \frac{1}{0.3}=\frac{3}{2}-1 \quad \frac{1}{R}-\frac{1}{-R}$
$$ \text { (Here, } R _1=R \text { and } R _2=-R \text { ) } $$
$$ R=0.3 $$
Now applying $\frac{\mu _2}{v}-\frac{\mu _1}{u}=\frac{\mu _2-\mu _1}{R}$ at air glass surface, we get
$$ \frac{3 / 2}{v _1}-\frac{1}{-(0.9)}=\frac{3 / 2-1}{0.3} $$
$$ \therefore \quad v _1=2.7 m $$
i.e. first image $I _1$ will be formed at $2.7 m$ from the lens. This will act as the virtual object for glass water surface. Therefore, applying $\frac{\mu _2}{v}-\frac{\mu _1}{u}=\frac{\mu _2-\mu _1}{R}$ at glass water surface, we have
$$ \begin{array}{rlrl} & \frac{4 / 3}{v _2}-\frac{3 / 2}{2.7} & =\frac{4 / 3-3 / 2}{-0.3} \\ \therefore \quad v _2 & =1.2 m \end{array} $$
i.e. second image $I _2$ is formed at $1.2 m$ from the lens or $0.4 m$ from the plane mirror. This will act as a virtual object for mirror. Therefore, third real image $I _3$ will be formed at a distance of $0.4 m$ in front of the mirror after reflection from it. Now this image will work as a real object for water-glass interface. Hence, applying
$$ \begin{aligned} \frac{\mu _2}{v}-\frac{\mu _1}{u} & =\frac{\mu _2-\mu _1}{R} \\ \text { we get } \frac{3 / 2}{v _4}-\frac{4 / 3}{-(0.8-0.4)} & =\frac{3 / 2-4 / 3}{0.3} \\ \therefore \quad v _4 & =-0.54 m \end{aligned} $$
i.e. fourth image is formed to the right of the lens at a distance of $0.54 m$ from it. Now finally applying the same formula for glass-air surface,
$$ \begin{array}{rlrl} \therefore \quad & \frac{1}{v _5}-\frac{3 / 2}{-0.54} & =\frac{1-3 / 2}{-0.3} \\ v _5 & =-0.9 m \end{array} $$
i.e. position of final image is $0.9 m$ relative to the lens (rightwards) or the image is formed $0.1 m$ behind the mirror.
