Question

A glass wedge of refractive index $\frac{3}{2}$ of inclination angle 0.01 radian is illuminated by monochromatic light of wavelength 6000 $\AA$ falling normally on it. If the $2^{nd}$ bright fringe formed by the reflected light is observed at a distance of x x $10^{-5}$ m from the edge of wedge, find x.

Answer 1

3

Answer 2

The problem describes interference in a wedge-shaped thin film.

1. Identify the given parameters:

• Refractive index of the glass wedge, $\mu = \frac{3}{2}$
• Inclination angle of the wedge, $\theta = 0.01$ radian
• Wavelength of monochromatic light, $\lambda = 6000 \AA = 6000 \times 10^{-10}$ m = $6 \times 10^{-7}$ m
• We need to find the distance 'x' from the edge of the wedge for the $2^{nd}$ bright fringe.

2. Understand phase change upon reflection: When light is incident normally on a thin film:

• The first reflection occurs at the air-glass interface. Since light reflects from a denser medium, a phase change of $\pi$ (or an equivalent path change of $\lambda/2$) occurs.
• The second reflection occurs at the glass-air interface. Since light reflects from a rarer medium, no phase change occurs.

Therefore, the net phase difference due to reflections between the two interfering rays is $\pi$, which corresponds to an additional path difference of $\lambda/2$.

3. Determine the condition for a bright fringe: The path difference due to the film thickness 't' is $2\mu t$ (for normal incidence).

Including the phase change due to reflection, the total effective path difference is $\Delta P = 2\mu t + \frac{\lambda}{2}$.

For constructive interference (bright fringe), the total effective path difference must be an integral multiple of the wavelength: $\Delta P = n\lambda$, where $n = 1, 2, 3, \dots$ (order of the bright fringe).

So, $2\mu t + \frac{\lambda}{2} = n\lambda$

$2\mu t = (n - \frac{1}{2})\lambda$

For the $2^{nd}$ bright fringe, $n=2$:

$2\mu t = (2 - \frac{1}{2})\lambda$

$2\mu t = \frac{3}{2}\lambda$

4. Relate film thickness 't' to distance 'x' for a wedge:

For a small wedge angle $\theta$, the thickness 't' at a distance 'x' from the edge is given by:

$t = x \tan\theta$

Since $\theta = 0.01$ radian is a small angle, $\tan\theta \approx \theta$.

So, $t = x\theta$.

5. Substitute and solve for 'x':

Substitute $t = x\theta$ into the bright fringe condition:

$2\mu (x\theta) = \frac{3}{2}\lambda$

Now, plug in the given values:

$2 \times \left(\frac{3}{2}\right) \times x \times (0.01) = \frac{3}{2} \times (6 \times 10^{-7} \text{ m})$

$3 \times x \times 0.01 = 3 \times 3 \times 10^{-7}$

$0.03x = 9 \times 10^{-7}$

$x = \frac{9 \times 10^{-7}}{0.03}$

$x = \frac{9 \times 10^{-7}}{3 \times 10^{-2}}$

$x = 3 \times 10^{-7 - (-2)}$

$x = 3 \times 10^{-5}$ m

The question asks for 'x' such that the distance is $x \times 10^{-5}$ m.

Comparing $3 \times 10^{-5}$ m with $X \times 10^{-5}$ m, we find $X = 3$.

The final answer is $\boxed{3}$.