A concave mirror of focal length 10 cm forms an image which is double the size of object when the object is placed at two different positions

Question

A concave mirror of focal length 10 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. [2026]

Smart Achievers · Accepted Answer

(10)

$Case-I$

$m = - 2 = \frac{- v}{u}$

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{2 u} + \frac{1}{u} = \frac{3}{2 u}$

$\Rightarrow u = \frac{3 f}{2} = \frac{3 \times (- 10)}{2} = - 15 cm$

$\Rightarrow u_{1} = - 15 cm$

$Case-II$

$m = + 2 = \frac{- v}{u}$

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{- 2 u} + \frac{1}{u} = \frac{- 1 + 2}{2 u} = \frac{1}{2 u}$

$\Rightarrow u_{2} = \frac{f}{2} = \frac{- 10}{2} = - 5 cm$

$d =$ $| u_{1} |$ $-$ $| u_{2} |$ $= 15 - 5 = 10 cm$

Solution

Q.
A concave mirror of focal length 10 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. [2026]

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Solution

Q. A concave mirror of focal length 10 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. [2026]

Ans. (10) Case-I m=-2=-vu 1f=1v+1u=12u+1u=32u ⇒u=3f2=3×(-10)2=-15 cm ⇒u1=-15 cm Case-II m=+2=-vu 1f=1v+1u=1-2u+1u=-1+22u=12u ⇒u2=f2=-102=-5 cm d=|u1|-|u2|=15-5=10 cm

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Q.
A concave mirror of focal length 10 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. [2026]