Optics 7 Question 46
47. The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature $20 cm$. The concave surface has a radius of curvature $60 cm$. The convex side is silvered and placed on a horizontal surface.
(1981, 2M)
(a) Where should a pin be placed on the optic axis such that its image is formed at the same place?
(b) If the concave part is filled with water of refractive index $4 / 3$, find the distance through which the pin should be moved, so that the image of the pin again coincides with the pin.
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Solution:
- (a) Image of object will coincide with it if ray of light after refraction from the concave surface fall normally on concave mirror so formed by silvering the convex surface. Or image after refraction from concave surface should form at centre of curvature of concave mirror or at a distance of $20 cm$ on same side of the combination. Let $x$ be the distance of pin from the given optical system.
Applying, $\frac{\mu _2}{v}-\frac{\mu _1}{u}=\frac{\mu _2-\mu _1}{R}$
With proper signs, $\frac{1.5}{-20}-\frac{1}{-x}=\frac{1.5-1}{-60}$
or $\frac{1}{x}=\frac{3}{40}-\frac{1}{120}=\frac{8}{120} \Rightarrow x=\frac{120}{8}=15 cm$
(b) Now, before striking with the concave surface, the ray is first refracted from a plane surface. So, let $x$ be the distance of pin, then the plane surface will form its image at a distance $\frac{4}{3} x\left(h _{\text {app. }}=\mu h\right)$ from it.
Now, using $\frac{\mu _2}{v}-\frac{\mu _1}{u}=\frac{\mu _2-\mu _1}{R}$ with proper signs,
we have $\frac{1.5}{-20}-\frac{4 / 3}{-\frac{4 x}{3}}=\frac{1.5-4 / 3}{-60}$
or $\quad \frac{1}{x}=\frac{3}{40}-\frac{1}{360}$ or $x=13.84 cm$
$\therefore \quad \Delta x=x _1-x _2=15 cm-13.84 cm$
$=1.16 cm$
(downwards)
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