A convex lens of focal length 20 cm produces images of the same magnification 2 when an object is kept at two distances $\mathrm{x}_{1}$ and $\mathrm{x}_{2}\left(\mathrm{x}_{1}>\mathrm{x}_{2}\right)$ from the lens. The ratio of $\mathrm{x}_{1}$ and $\mathrm{x}_{2}$ is:

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Geometrical Optics

A convex lens of focal length 20 cm produces images of the same magnification 2 when an object is kept at two distances

x_{1}

and

x_{2} (x_{1} > x_{2})

from the lens. The ratio of

x_{1}

and

x_{2}

is:

(A) $5 : 3$
(B) $2 : 1$
(C) $4 : 3$
(D) $3 : 1$ ✓ Correct

Answer: (D)

Solution

Keeping the magnitude of magnification at

2

, a convex lens of focal length

20

cm can form either a real inverted image or a virtual erect one. Writing the magnification as

m = \frac{f}{f + u} = \frac{20}{20 - x}

for an object at distance

x

, the real-image case needs

m = - 2

\frac{20}{20 - x} = - 2

gives

x_{1} = 30

cm. The virtual-image case needs

m = + 2

\frac{20}{20 - x} = 2

gives

x_{2} = 10

cm. The larger distance to the smaller one is

\frac{x _{1}}{x _{2}} = \frac{30}{10} = 3 : 1

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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