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

Step 1: Understanding the Concept:
For refraction at a single spherical boundary, the image position can be located using the spherical refraction formula. Once the image distance is found, the transverse (linear) magnification is calculated using its specific formula.
Step 2: Key Formula or Approach:
Spherical refraction formula: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \] Transverse magnification formula: \[ m = \left(\frac{n_1}{n_2}\right) \frac{v}{u} \]
Step 3: Detailed Explanation:
Using the standard sign convention:
Object distance \(u = -4R\).
Radius of curvature is \(+R\) (convex boundary towards rarer medium).
Refractive indices: \(n_1 = 1\), \(n_2 = 1.4\).
Substitute into the refraction formula:
\[ \frac{1.4}{v} - \frac{1}{-4R} = \frac{1.4 - 1}{R} \] \[ \frac{1.4}{v} + \frac{1}{4R} = \frac{0.4}{R} \] \[ \frac{1.4}{v} = \frac{0.4}{R} - \frac{0.25}{R} = \frac{0.15}{R} \] Solve for \(v\): \[ v = \frac{1.4R}{0.15} = \frac{140R}{15} = \frac{28R}{3} \] Now, substitute \(v\) and \(u\) into the magnification formula: \[ m = \left(\frac{1}{1.4}\right) \frac{\left(\frac{28R}{3}\right)}{-4R} \] \[ m = \left(\frac{1}{1.4}\right) \left(-\frac{7}{3}\right) = -\frac{7}{4.2} = -\frac{70}{42} \] Simplify the fraction: \[ m = -\frac{5}{3} \approx -1.67 \]
Step 4: Final Answer:
The transverse magnification is approximately \(–1.67\).