JEE PYQ: Ray Optics Question 44
Question 44 - 2019 (09 Jan Shift 1)
A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance $d$. Then $d$ is:
(1) 1.1 cm away from the lens
(2) 0
(3) 0.55 cm towards the lens
(4) 0.55 cm away from the lens
Show Answer
Answer: (4)
Solution
Using lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\frac{1}{10} - \frac{1}{-10} = \frac{1}{f} \Rightarrow f = 5$ cm Shift due to slab $= t\left(1 - \frac{1}{\mu}\right)$ in the direction of incident ray or, $d = 1.5\left(1 - \frac{2}{3}\right) = 0.5$ Now, $u = -9.5$ Again using lens formula $\frac{1}{v} - \frac{1}{-9.5} = \frac{1}{5}$ $\Rightarrow \frac{1}{v} = \frac{1}{5} - \frac{2}{19} = \frac{9}{95}$ or, $v = \frac{95}{9} = 10.55$ cm Thus, screen is shifted by a distance $d = 10.55 - 10 = 0.55$ cm away from the lens.
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