Variations in YDSE (Young's Double Slit Experiment) Questions in English

(C) The path difference between the waves reaching point $P$ from $S_1$ and $S_2$ is given by $\Delta x = S_1 P - S_2 P$.
For the first bright fringe,the path difference must be equal to the wavelength,so $\Delta x = \lambda$.
From the geometry of the setup,the path difference can also be expressed as $\Delta x = d \cos \theta$,where $d = 2 \lambda$ is the separation between the sources.
Thus,$2 \lambda \cos \theta = \lambda$.
This simplifies to $\cos \theta = \frac{1}{2}$,which means $\theta = 60^{\circ}$.
From the right-angled triangle formed by the screen,we have $\tan \theta = \frac{OP}{D}$.
Substituting $\theta = 60^{\circ}$,we get $\tan 60^{\circ} = \frac{OP}{D}$.
Since $\tan 60^{\circ} = \sqrt{3}$,we have $\sqrt{3} = \frac{OP}{D}$.
Therefore,$OP = \sqrt{3} D$.

(A) When a thin sheet of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits, the path difference introduced is $\Delta x = (\mu - 1)t$.
Given that the central point on the screen is now occupied by the $7^{th}$ bright fringe, the shift in the fringe pattern is equal to $7$ fringe widths.
The condition for the shift is $(\mu - 1)t = n\lambda$, where $n = 7$ and $\lambda = 600 \ nm = 0.6 \ \mu m$.
Substituting the values: $(1.6 - 1)t = 7 \times 0.6 \ \mu m$.
$0.6 \times t = 4.2 \ \mu m$.
$t = \frac{4.2}{0.6} \ \mu m = 7 \ \mu m$.

(C) The shift in the central bright fringe in a Young's double slit experiment when a thin transparent sheet is introduced is given by the formula: $y_{\text{shift}} = \frac{(\mu - 1) t D}{d}$.
Given values are: $d = 0.4 \ mm = 0.4 \times 10^{-3} \ m$,$D = 1 \ m$,$t = 20 \ \mu m = 20 \times 10^{-6} \ m$,and $y_{\text{shift}} = 20 \ mm = 20 \times 10^{-3} \ m$.
Substituting these values into the formula:
$20 \times 10^{-3} = \frac{(\mu - 1) \times 20 \times 10^{-6} \times 1}{0.4 \times 10^{-3}}$
$20 \times 10^{-3} = \frac{(\mu - 1) \times 20 \times 10^{-6}}{0.4 \times 10^{-3}}$
$20 \times 10^{-3} = (\mu - 1) \times 50 \times 10^{-3}$
$\mu - 1 = \frac{20}{50} = 0.4$
$\mu = 1.4$
Since the refractive index is given as $\frac{\alpha}{10} = 1.4$,we get $\alpha = 14$.

(A) The shift in the central fringe in a Young's Double Slit Experiment $(YDSE)$ when a transparent sheet of thickness $t$ and refractive index $\mu$ is introduced is given by the formula: $\Delta x = \frac{D}{d}(\mu - 1)t$.
Given values are: $d = 0.1 \ cm$,$D = 50 \ cm$,$\Delta x = 0.2 \ cm$,and $\mu = 1.5$.
Rearranging the formula to solve for $t$: $t = \frac{\Delta x \cdot d}{D(\mu - 1)}$.
Substituting the values: $t = \frac{0.2 \times 0.1}{50(1.5 - 1)}$.
$t = \frac{0.02}{50 \times 0.5} = \frac{0.02}{25}$.
$t = 0.0008 \ cm = 8 \times 10^{-4} \ cm$.

(A) The shift in the central fringe due to the introduction of a mica sheet is given by $\Delta y = \frac{(\mu - 1)tD}{d}$.
Given that this shift is equal to the position of the $7^{th}$ bright fringe,we have $\Delta y = 7 \times \frac{\lambda D}{d}$.
Equating the two expressions: $\frac{(\mu - 1)tD}{d} = \frac{7\lambda D}{d}$.
This simplifies to $(\mu - 1)t = 7\lambda$.
Substituting the given values: $(1.56 - 1)t = 7 \times 450 \times 10^{-9} \text{ m}$.
$0.56t = 3150 \times 10^{-9} \text{ m}$.
$t = \frac{3150}{0.56} \times 10^{-9} \text{ m} = 5625 \times 10^{-9} \text{ m}$.
Comparing this with $\alpha \times 10^{-9} \text{ m}$,we get $\alpha = 5625$.