First slide

The lens

Question

In the given figure, the image of object AB observed by observer is

Difficult

A

virtual and erect
B

1.5 cm in length
C

30 cm from the lens
D

60 cm from the lens

Solution

Object shiftness $x = t (1 - \frac{1}{μ}) = 15 (1 - \frac{2}{3}) = 5 c m$
The image of AB is $A_{1} B_{1}$ due to refraction through slab, which is shifted 5 cm from AB. The size of $A_{1} B_{1}$ is same as that of AB.
$A_{1} B_{1}$ Behaves as object for the lens.
For convex lens, $A_{1} B_{1}$ behaves as object.
For lens, $u = - 60 c m, f = 20 c m$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{v} + \frac{1}{60} = \frac{1}{20}$
Or    $\frac{1}{v} = \frac{1}{20} - \frac{1}{60} = \frac{3 - 1}{60}$
$∴ v = + 30 c m$
$∵ \frac{A_{2} B_{2}}{A_{1} B_{1}} = \frac{v}{u}$
Or $A_{2} B_{2} = \frac{v}{u} A_{1} B_{1} = (\frac{30}{- 60}) 2 c m = - 1 c m$
Thus, final image $A_{2} B_{2}$ is real, inverted, 1 cm in length and is 30 cm from the lens.

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