Ray Optics

JEE Main level, numerical answer, medium

18 of 295

Question 18 of 295

A concave mirror of focal length 1010 cm forms an image which is double the size of object when the object is placed at two different positions. The distance between the two positions of the object is __________ cm.

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Answer: 10

Given focal length of the concave mirror, f=−10f = -10 cm.

The image is double the size of the object, so the magnification can be m=±2m = \pm 2.

Using the magnification formula in terms of focal length and object distance:

m=ff−um = \dfrac{f}{f - u}

Case 1: For a real image, m=−2m = -2

−2=−10−10−u1-2 = \dfrac{-10}{-10 - u_1}

20+2u1=−1020 + 2u_1 = -10

2u1=−30⇒u1=−152u_1 = -30 \Rightarrow u_1 = -15 cm

Case 2: For a virtual image, m=+2m = +2

2=−10−10−u22 = \dfrac{-10}{-10 - u_2}

−20−2u2=−10-20 - 2u_2 = -10

−2u2=10⇒u2=−5-2u_2 = 10 \Rightarrow u_2 = -5 cm

The distance between the two positions of the object is:

Δu=∣u1−u2∣=∣−15−(−5)∣=10\Delta u = |u_1 - u_2| = |-15 - (-5)| = 10 cm

Answer: 1010

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