We are asked to find the percentage change in the fringe width when the slit separation in a Young’s double slit experiment is increased from \( 0.2 \, \mathrm{mm} \) to \( 0.4 \, \mathrm{mm} \).
The fringe width \( \beta \) in a Young’s double slit experiment is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
where:
\( \lambda \) = wavelength of light used
\( D \) = distance between slits and the screen
\( d \) = separation between the two slits
Hence, fringe width \( \beta \) is inversely proportional to slit separation \( d \):
\[ \beta \propto \frac{1}{d} \]
Step 1: Let initial slit separation \( d_1 = 0.2 \, \mathrm{mm} \), and final slit separation \( d_2 = 0.4 \, \mathrm{mm} \).
Step 2: Write the ratio of fringe widths before and after the change.
\[ \frac{\beta_2}{\beta_1} = \frac{d_1}{d_2} \]
Step 3: Substitute the given values.
\[ \frac{\beta_2}{\beta_1} = \frac{0.2}{0.4} = \frac{1}{2} \]
Step 4: Hence, the new fringe width is half of the original fringe width.
\[ \beta_2 = \frac{1}{2} \beta_1 \]
Step 5: Calculate the percentage change in fringe width.
\[ \text{Percentage change} = \frac{\beta_2 - \beta_1}{\beta_1} \times 100 = \frac{\frac{1}{2}\beta_1 - \beta_1}{\beta_1} \times 100 \] \[ = (-0.5) \times 100 = -50\% \]
The negative sign indicates a decrease in fringe width.
Final Answer: The fringe width decreases by 50%.
\[ \boxed{\text{Percentage change in fringe width = } -50\%} \]
If $ \theta \in [-2\pi,\ 2\pi] $, then the number of solutions of $$ 2\sqrt{2} \cos^2\theta + (2 - \sqrt{6}) \cos\theta - \sqrt{3} = 0 $$ is:
A thin transparent film with refractive index 1.4 is held on a circular ring of radius 1.8 cm. The fluid in the film evaporates such that transmission through the film at wavelength 560 nm goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, the rate of evaporation is:
The major product (A) formed in the following reaction sequence is
