SATHEE: Optics 7 Question 41

Optics 7 Question 41

42. A thin equiconvex lens of glass of refractive index $μ = 3 / 2$ and of focal length $0.3 m$ in air is sealed into an opening at one end of a tank filled with water $μ = 4 / 3$ . 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,

$\frac{1}{f} = (μ - 1) \frac{1}{R_{1}} - \frac{1}{R_{2}}$

we have $\frac{1}{0.3} = \frac{3}{2} - 1 \frac{1}{R} - \frac{1}{- R}$

$(Here, R_{1} = R and R_{2} = - R)$

$R = 0.3$

Now applying $\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$ at air glass surface, we get

$\frac{3 / 2}{v_{1}} - \frac{1}{- (0.9)} = \frac{3 / 2 - 1}{0.3}$

$∴ v_{1} = 2.7 m$

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 $\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$ at glass water surface, we have

$\begin{aligned} \frac{4 / 3}{v_{2}} - \frac{3 / 2}{2.7} & = \frac{4 / 3 - 3 / 2}{- 0.3} \\ ∴ v_{2} & = 1.2 m \end{aligned}$

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

$\begin{aligned} \frac{μ_{2}}{v} - \frac{μ_{1}}{u} & = \frac{μ_{2} - μ_{1}}{R} \\ we get \frac{3 / 2}{v_{4}} - \frac{4 / 3}{- (0.8 - 0.4)} & = \frac{3 / 2 - 4 / 3}{0.3} \\ ∴ v_{4} & = - 0.54 m \end{aligned}$