Question

Find transverse magnification due to curved boundary between two mediums as shown in the figure. Refractive index of second medium \(n_2 = 1.4\).

• A. \(-1.5\)
• B. \(-1.67\)
• C. \(+1.2\)
• D. \(-0.8\)

Answer 1

The Correct Option is B

Concept: For refraction at a spherical surface, \[ \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \] Transverse magnification is \[ m=\frac{n_1 v}{n_2 u} \] Step 1: Apply refraction formula at spherical surface.} Given: \[ n_1=1, \quad n_2=1.4, \quad u=-4R \] \[ \frac{1.4}{v}-\frac{1}{-4R}=\frac{1.4-1}{R} \] \[ \frac{1.4}{v}+\frac{1}{4R}=\frac{0.4}{R} \]
Step 2: Solve for \(v\).} \[ \frac{1.4}{v}=\frac{0.4}{R}-\frac{1}{4R} \] \[ \frac{1.4}{v}=\frac{4}{10R}-\frac{1}{4R} \] \[ \frac{1.4}{v}=\frac{6}{40R} \] \[ v=\frac{56R}{6}=\frac{28}{3}R \]
Step 3: Find transverse magnification.} \[ m=\frac{n_1 v}{n_2 u} \] \[ m=\frac{1 \times \frac{28R}{3}}{1.4 \times (-4R)} \] \[ m=-1.67 \] Final Result \[ m=-1.67 \]