Q.50. A concave mirror of focal length 10 cm forms an image which is double the size of object when

Q: What is the correct answer for this exam Physics Q50?

The correct answer is A. Focal length of concave mirror $f = -10 cm$. Magnification $m = -v/u$. Given image is double the size of the object, so $|m| = 2$. Case 1: Real image. $m = -2$. So $-v/u = -2 \Rightarrow v = 2u$. Using mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. $\frac{1}{-10} = \frac{1}{2u} + \frac{1}{u} = \frac{1+2}{2u} = \frac{3}{2u}$. $2u = -30 \Rightarrow u_1 = -15 cm$. Case 2: Virtual image. $m

Q: Q.50. 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…

The correct answer is A. Focal length of concave mirror $f = -10 cm$. Magnification $m = -v/u$. Given image is double the size of the object, so $|m| = 2$. Case 1: Real image. $m = -2$. So $-v/u = -2 \Rightarrow v = 2u$. Using mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. $\frac{1}{-10} = \frac{1}{2u} + \frac{1}{u} = \frac{1+2}{2u} = \frac{3}{2u}$. $2u = -30 \Rightarrow u_1 = -15 cm$. Case 2: Virtual image. $m

Q: What are the options for this this exam Physics question?

A. 10 · B. 20 · C. 15 · D. 5

Question

Q.50. 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.

FirstInTest · Accepted Answer

Focal length of concave mirror $f = -10 cm$. Magnification $m = -v/u$. Given image is double the size of the object, so $|m| = 2$. Case 1: Real image. $m = -2$. So $-v/u = -2 \Rightarrow v = 2u$. Using mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. $\frac{1}{-10} = \frac{1}{2u} + \frac{1}{u} = \frac{1+2}{2u} = \frac{3}{2u}$. $2u = -30 \Rightarrow u_1 = -15 cm$. Case 2: Virtual image. $m = +2$. So $-v/u = +2 \Rightarrow v = -2u$. Using mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. $\frac{1}{-10} = \frac{1}{-2u} + \frac{1}{u} = \frac{-1+2}{2u} = \frac{1}{2u}$. $2u = -10 \Rightarrow u_2 = -5 cm$. The distance between the two positions of the object is $|u_1 - u_2| = |-15 - (-5)| = |-15 + 5| = |-10| = 10 cm$.

Answer

10

Answer

20

Answer

15

Answer

5

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