Question

For the following question, enter the correct numerical value upto TWO decimal places. If the numerical value has more than two decimal places, round-off the value to TWO decimal places. (For example: Numeric value 5 will be written as 5.00 and 2.346 will be written as 2.35) The image of an object placed in front of a concave mirror of focal length 12 cm is formed at a point which is 10 cm more distant from the mirror than the object. The magnification of the image is ____.

Hint:

For mirrors, always apply the sign convention carefully. Magnification \(m\) is negative for real, inverted images formed by concave mirrors.

Answer 1

Correct Answer: 1.5

Step 1: Let the object distance from the mirror be \(u\) cm and the image distance be \(v\) cm. Given that the image is formed 10 cm farther from the mirror than the object: \[ v = u + 10 \]
Step 2: Using the mirror formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] For a concave mirror: \[ f = -12 \text{ cm} \] Substitute: \[ \frac{1}{-12} = \frac{1}{u+10} + \frac{1}{u} \]
Step 3: Simplifying: \[ -\frac{1}{12} = \frac{2u+10}{u(u+10)} \] \[ u(u+10) = -12(2u+10) \] \[ u^2 + 10u = -24u - 120 \] \[ u^2 + 34u + 120 = 0 \] Solving: \[ u = -20 \text{ cm} \]
Step 4: Then, \[ v = u + 10 = -20 + 10 = -30 \text{ cm} \]
Step 5: Magnification for a mirror is: \[ m = -\frac{v}{u} \] \[ m = -\frac{-30}{-20} = -1.5 \] Final Answer (up to two decimal places): \[ \boxed{-1.50} \]