Question

A spherical mirror has principal axis along x-axis. A real object 'O' is placed on x axis. Image of the object is formed real with magnification $-\frac{1}{3}$. The magnitude of difference between x-coordinates of the object and image is 10 cm. Find magnitude of 4f.

• A. 15 cm
• B. 30 cm
• C. 45 cm
• D. 60 cm

Answer 1

Answer: (B)

30 cm

Answer 2

Let the pole of the mirror be at the origin. Given magnification $m = -1/3$. For axial points, $m = -x_I/x_O$. Thus, $-1/3 = -x_I/x_O$, which implies $x_O = 3x_I$. We are given $|x_O - x_I| = 10$ cm. Substituting $x_O = 3x_I$, we get $|3x_I - x_I| = 10$, so $|2x_I| = 10$, which means $|x_I| = 5$ cm. Since the image is real and formed by a real object, $x_I$ and $x_O$ must have the same sign. Assuming the object is placed to the left of the pole (real object convention), $x_O < 0$, so $x_I < 0$. Thus, $x_I = -5$ cm and $x_O = 3 \times (-5) = -15$ cm. Using the mirror formula $\frac{1}{f} = \frac{1}{x_I} - \frac{1}{x_O}$, we get $\frac{1}{f} = \frac{1}{-5} - \frac{1}{-15} = -\frac{1}{5} + \frac{1}{15} = \frac{-3+1}{15} = -\frac{2}{15}$. Therefore, $f = -15/2 = -7.5$ cm. The magnitude of $4f$ is $|4 \times (-7.5)| = |-30| = 30$ cm.