Suppose a pin is placed in front of a concave mirror and a real image that is thrice the size of the pin is formed on a screen. The pin and the screen are then moved until the image is six times the size of the object. If the shift of the screen is 24 cm, then the shift in the object is:

Correct option is B

Solution:

Given:

• A concave mirror forms a real image that is thrice the size of the pin initially.
• The image is then six times the size of the pin after shifting the object and the screen.
• The shift of the screen is 24 cm.

​$$M = -\frac{v}{u}$$​​

where:

• v is the image distance (positive for real images),
• u is the object distance (negative for objects in front of the mirror).

Also, the mirror equation relates object distance (uuu), image distance (v), and the focal length (f):

​$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$​​

Formula Used:

1. Magnification: $$M = -\frac{v}{u}$$​
2. Mirror equation:$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$​

Solution:

Substitute $$\frac{u_2}{u_1} = \frac{7}{8}$$​​ into Equation 4:

​$$6\left(\frac{7}{8}u_1\right) - 3u_1 = 24$$​​

Simplify:

​$$\frac{42}{8}u_1 - 3u_1 = 24 \implies \frac{18}{8}u_1 = 24\\ \ \\u_1 = \frac{24 \times 8}{18} = \frac{192}{18} \, \text{cm}$$​​

The shift in the object is:

$$\Delta u = u_2 - u_1 = \frac{7}{8}u_1 - u_1 = \left(\frac{7}{8} - 1\right)u_1 = \frac{-1}{8}\\ \ \\\Delta u = \frac{-1}{8} \times \frac{192}{18} = -\frac{4}{3} \, \text{cm}$$​

The shift in the object is 1.33 cm towards the mirror.