A concave mirror forms a real image of three times the size of an object on a screen. The object and screen are then moved until the image is six times the size of the object. If the shift of the screen is 39 cm then the focal length of the mirror is:

Correct option is C

Correct Answer: (c). 13 cm

Solution:

1. Given:
• Magnifications:$$M_1 = 3, M_2 = 6$$​,
• Screen shift: Δv=39 cm
2. Key equations:
• Image distances:$$v_1 = -3u_1​, \ \ v_2 = -6u_2​,$$​​​
• Shift:$$v_2 - v_1 = 39 \implies -6u_2 + 3u_1 = 39$$​​
• Focal length:$$f = \frac{3u_1}{2} = \frac{6u_2}{5}$$​​​.
3. Solve:
• From$$\frac{3u_1}{2} = \frac{6u_2}{5},u_1 = \frac{4u_2}{5}$$​,
• Substitute into$$-6u_2 + 3u_1 = 39$$​:$$.-6u_2 + 3\left(\frac{4u_2}{5}\right) = 39 \implies u_2 = -10.83 \, \text{cm}, \, u_1 = -8.67$$​​
4. Focal length:
• ​$$\frac{3u_1}{2} = -13 \, \text{cm}$$​​
  

Final Answer:

The focal length is 13 cm.