Question

In Y.D.S.E. the fringe width is 0.2 mm. If wavelength of light is increase by 10% and separation between the slits is increased by 10% then fringe width will be :

• A. 0.20 mm
• B. 0.165 mm
• C. 0.401 mm
• D. 0.242 mm

Answer 1

Answer: (A)

0.20 mm

Answer 2

The fringe width ($\beta$) in Young's Double Slit Experiment (YDSE) is given by the formula:

$$\beta = \frac{\lambda D}{d}$$

where:

• $\lambda$ is the wavelength of the light used.
• $D$ is the distance between the slits and the screen.
• $d$ is the separation between the two slits.

Given:

Initial fringe width, $\beta_1 = 0.2$ mm.

The wavelength of light is increased by 10%. So, the new wavelength ($\lambda_2$) will be:

$$\lambda_2 = \lambda_1 + 0.10 \lambda_1 = 1.10 \lambda_1$$

The separation between the slits is increased by 10%. So, the new slit separation ($d_2$) will be:

$$d_2 = d_1 + 0.10 d_1 = 1.10 d_1$$

The distance between the slits and the screen ($D$) is not mentioned to be changed, so we assume it remains constant, i.e., $D_2 = D_1$.

Now, let's write the expression for the new fringe width ($\beta_2$):

$$\beta_2 = \frac{\lambda_2 D_2}{d_2}$$

Substitute the new values of $\lambda_2$, $D_2$, and $d_2$:

$$\beta_2 = \frac{(1.10 \lambda_1) D_1}{(1.10 d_1)}$$

We can rearrange the terms:

$$\beta_2 = \left( \frac{1.10}{1.10} \right) \times \left( \frac{\lambda_1 D_1}{d_1} \right)$$

Since $\frac{1.10}{1.10} = 1$, and we know that $\beta_1 = \frac{\lambda_1 D_1}{d_1}$, the equation becomes:

$$\beta_2 = 1 \times \beta_1$$ $$\beta_2 = \beta_1$$

Given $\beta_1 = 0.2$ mm.

Therefore, the new fringe width $\beta_2 = 0.2$ mm.

Explanation:

The fringe width in YDSE is directly proportional to the wavelength ($\lambda$) and inversely proportional to the slit separation ($d$). When both the wavelength and the slit separation are increased by the same percentage (10% in this case), the factor of increase cancels out.

New fringe width $\beta' = \frac{(1.1 \lambda) D}{(1.1 d)} = \frac{1.1}{1.1} \frac{\lambda D}{d} = \frac{\lambda D}{d} = \beta_{\text{initial}}$.

Thus, the fringe width remains unchanged.