JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Geometrical Optics

Distance between object and its image (magnified by

- \frac{1}{3}

) is 30 cm. The focal length of the mirror used is

(\frac{x}{4})

cm, where magnitude of value of

x

is ______.

Answer: 45

Solution

For a mirror

m = - v / u

, and

m = - \frac{1}{3}

gives

v = u /3

, a real image on the same side as the object. Their separation is

∣ u ∣ - ∣ v ∣ = ∣ u ∣ - \frac{∣ u ∣}{3} = \frac{2}{3} ∣ u ∣ = 30 cm \Rightarrow ∣ u ∣ = 45 cm, ∣ v ∣ = 15 cm .

From the mirror formula,

\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = - \frac{1}{15} - \frac{1}{45} = - \frac{4}{45} \Rightarrow f = - \frac{45}{4} cm .

Writing

f = x /4

, the magnitude gives

x = 45

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Integer answer

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