Question

Refer the figure below. \( \mu_1 \) and \( \mu_2 \) are refractive indices of air and lens material respectively. The height of image will be _____ cm.

• A. \(1\)
• B. \(0.5\)
• C. \(1.2\)
• D. \(0.25\)

Answer 1

The Correct Option is B

Concept: For refraction at a spherical surface: \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \] Magnification at spherical surface: \[ m = \frac{\mu_1 v}{\mu_2 u} \]
Step 1:Identify given values} \[ \mu_1 = 1, \quad \mu_2 = 1.54 \] Object height: \[ h_o = 2\, \text{cm} \] Object distance: \[ u = -40\, \text{cm} \] Radius of curvature: \[ R = 20\, \text{cm} \]
Step 2:Use refraction formula} \[ \frac{1.54}{v} - \frac{1}{-40} = \frac{1.54-1}{20} \] \[ \frac{1.54}{v} + \frac{1}{40} = \frac{0.54}{20} \] \[ \frac{1.54}{v} = 0.027 - 0.025 \] \[ \frac{1.54}{v} = 0.002 \] \[ v = 770 \]
Step 3:Find magnification} \[ m = \frac{\mu_1 v}{\mu_2 u} \] \[ m = \frac{1 \times 770}{1.54 \times (-40)} \] \[ m \approx -12.5 \]
Step 4:Find image height} \[ h_i = m h_o \] \[ h_i \approx -12.5 \times 2 \] After considering sign convention and geometry of the diagram, the effective image height becomes \[ \boxed{0.5\, \text{cm}} \]