Question

A collimated beam of laser light of wavelength 514 nm is normally incident on a smooth glass slab placed in air. Given the refractive indices of glass and air are 1.47 and 1.0 respectively, the percentage of light intensity reflected back is:

• A. 0
• B. 4.0
• C. 3.6
• D. 4.2

Hint:

The greater the difference in refractive indices between two media, the higher the reflection at the boundary.

Answer 1

The Correct Option is C

Step 1: Formula for reflection coefficient.
The reflection coefficient for normal incidence is \[ R = \left( \frac{n_1 - n_2}{n_1 + n_2} \right)^2 \] where \( n_1 = 1.47 \) (glass) and \( n_2 = 1.0 \) (air).

Step 2: Substituting values.
\[ R = \left( \frac{1.47 - 1.0}{1.47 + 1.0} \right)^2 = \left( \frac{0.47}{2.47} \right)^2 = 0.036 \] Hence, reflected percentage \( = 3.6%. \)

Step 3: Conclusion.
The percentage of reflected light intensity is \(3.6%\).