To find the width of each slit that results in 10 maxima of the double-slit pattern within the central maximum of the single-slit pattern, we begin by analyzing the interference and diffraction conditions.
For a double-slit, the condition for maxima is:
\(d \sin \theta = m\lambda\),
where \(d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m}\) (distance between slits), \(\lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m}\), and \(m\) is the order of the maximum.
The angular width of the central maximum of single-slit diffraction is given by:
\(\sin \theta = \frac{\lambda}{a}\),
where \(a\) is the slit width.
Given that 10 maxima of the double-slit pattern fit within the central maximum of the single-slit pattern, the angle for the 5th order (half of 10) double-slit maximum should equal the angle for the first minimum of the single-slit pattern:
\(\frac{5\lambda}{d} = \frac{\lambda}{a}\).
Solving for \(a\):
\(a = \frac{d}{5} = \frac{1 \times 10^{-3}}{5} = 0.2 \times 10^{-3} \, \text{m} = 2 \times 10^{-4} \, \text{m}\).
This calculated width of the slit \(a = 2 \times 10^{-4} \, \text{m}\) falls within the specified range. Therefore, the width of each slit is \(2 \times 10^{-4} \, \text{m}\).
Given:
\[d = 1 \, \text{mm} = 10^{-3} \, \text{m}, \quad D = 1 \, \text{m}, \quad \lambda = 500 \, \text{nm} = 5 \times 10^{-7} \, \text{m}.\]
For the central maximum of the single-slit pattern to contain 10 maxima of the double-slit pattern:
\[10 \times \frac{\lambda D}{d} = \frac{2\lambda D}{a},\]
where $a$ is the slit width.
Rearranging for $a$:
\[a = \frac{d}{5}.\]
Substitute $d = 10^{-3} \, \text{m}$:
\[a = \frac{10^{-3}}{5} = 2 \times 10^{-4} \, \text{m}.\]
Thus, the slit width is:
\[a = 2 \times 10^{-4} \, \text{m}.\]
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)