Question

A thin film of refractive index 1.5 and thickness `4 xx 10^(-5)` cm is illuminated by light normal to the surface. What wavelength within the visible spectrum will be intensified in the reflected beam?
A. `4800 Å`
B. `5800 Å`
C. `6000 Å`
D. `6800 Å`

Answer 1

Correct Answer - a
Condition for observing bright fringe is
`2 nd = (m + (1)/(2)) lambda`
`:. lambda = (2 nd)/((m + (1)/(2))) = (2 xx 1.5 xx 4 xx 10^(-5))/(m + (1)/(2))`
`= (12 xx 10^(-5))/(m + (1)/(2))`
The integer m that gives the wavelength in the visible region (`4000 Å` to `7000 Å`) is `m = 2`. In that case,
`lambda = (12 xx 10^(-5))/(2 (1)/(2)) = 4.8 xx 10^(-5) = 4800 Å`