Given:
Focal length, \( f = 15 \, \text{cm} \)
Object distance, \( u = -30 \, \text{cm} \) (negative for real object)
Step 1: Mirror Formula The mirror formula is given by: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] where: - \( f \) is the focal length, - \( v \) is the image distance, - \( u \) is the object distance.
Step 2: Solve for Image Distance \( v \) Substitute the given values into the mirror formula: \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-30} \] \[ \frac{1}{15} = \frac{1}{v} - \frac{1}{30} \] \[ \frac{1}{v} = \frac{1}{15} + \frac{1}{30} = \frac{2 + 1}{30} = \frac{3}{30} = \frac{1}{10} \] \[ v = 10 \, \text{cm} \]
Answer: The correct answer is option (c): 60 cm.
A transparent block A having refractive index $ \mu_2 = 1.25 $ is surrounded by another medium of refractive index $ \mu_1 = 1.0 $ as shown in figure. A light ray is incident on the flat face of the block with incident angle $ \theta $ as shown in figure. What is the maximum value of $ \theta $ for which light suffers total internal reflection at the top surface of the block ?