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%.
\[ \boxed{\text{Percentage change in fringe width = } -50\%} \]
Nature of compounds TeO₂ and TeH₂ is___________ and ______________respectively.
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
The magnitude of heat exchanged by a system for the given cyclic process ABC (as shown in the figure) is (in SI units):
