Wave Optics

In Young's double slit experiment.

Angular fringe width, $\theta=0.15^{\circ}$

$ D=80 \mathrm{~cm}=0.8 \mathrm{~m} $

Wavelength of light used,

$ \lambda=490 \mathrm{~nm}=4.9 \times 10^{-7} \mathrm{~m} $

Angular fringe width when entire apparatus is immersed in a medium of refractive index $\mu=5 / 3$, is given as,

$ \theta^{\prime}=\frac{\theta}{\mu} $

$ =\frac{0.15}{5 / 3}=\frac{0.15 \times 3}{5}=0.09^{\circ} $

Given, wavelength of red light,

$ \lambda_R=780 \mathrm{~nm}=7.8 \times 10^{-7} \mathrm{~m} $

Wavelength of blue light,

$ \lambda_B=520 \mathrm{~nm}=5.2 \times 10^{-7} \mathrm{~m} $

According to question,

$ \begin{aligned} & & x_R & =x_B \\ \Rightarrow & & \frac{n \lambda_R D}{d} & =\frac{(n+1) \lambda_B D}{d} \end{aligned} $

$ \Rightarrow \quad n \lambda_R=(n+1) \lambda_B \Rightarrow \frac{\lambda_R}{\lambda_B}=\frac{n+1}{n} $

$ \begin{aligned} \Rightarrow & & \frac{7.8 \times 10^{-7}}{52 \times 10^{-7}} & =\frac{n+1}{n} \Rightarrow \frac{39}{26}=\frac{n+1}{n} \\ \Rightarrow & & 39 n & =26 n+26 \\ \Rightarrow & & 39 n-26 n & =26 \\ \Rightarrow & & 13 n & =26 \Rightarrow n=2 \end{aligned} $

Given, when young's double slit experiment is performed in air, the angular fringe width is $\theta=0.2$

$ =0.2 \times \frac{\pi}{180} \mathrm{rad} $

But we know that,

$ \theta=\frac{\lambda}{d} $

$ \begin{aligned} \Rightarrow \quad d & =\frac{\lambda}{\theta}=\frac{600 \times 10^{-9}}{0.2 \times \frac{\pi}{180}} \\ & {\left[\because \lambda=600 \mathrm{~nm}=600 \times 10^{-9} \mathrm{~m}\right] } \\ & =\frac{600 \times 180 \times 10^{-9}}{0.2 \times \pi} \\ & =\frac{1.08 \times 10^{-4}}{0.2 \pi} \mathrm{~m} \end{aligned} $

When the entire measurement system is immersed in water of refractive index $\mu=\frac{4}{3}$, then new value of angular fringe width.

$ \begin{aligned} \theta^{\prime} & =\frac{\lambda}{d \mu}=\frac{600 \times 10^{-9}}{\frac{1.08 \times 10^{-4}}{0.2 \pi} \times \frac{4}{3}} \\ & =\frac{3 \times 0.2 \pi \times 600 \times 10^{-9}}{1.08 \times 4 \times 10^{-4}} \\ & =2.62 \times 10^{-3} \mathrm{rad} \\ & =2.62 \times 10^{-3} \times \frac{180}{\pi} \text { degree } \\ & =015 \text { degree } \end{aligned} $

∴ Change in angular width $=\theta-\theta^{\prime}$

$ =0.2^{\circ}-0.15^{\circ}=0.05^{\circ} $

In Young's double slit experiment,

$ \begin{aligned} & d=0.2 \mathrm{~mm}=2 \times 10^{-4} \mathrm{~m} \\ & D=60 \mathrm{~cm}=0.6 \mathrm{~m} \\ & \lambda=550 \mathrm{~nm}=5.5 \times 10^{-7} \mathrm{~m} \\ & \mu=\frac{11}{9} \end{aligned} $

∴ Fringe width when whole apparatus is immersed in liquid,

$ \begin{aligned} \beta & =\frac{D \lambda}{d \mu}=\frac{0.6 \times 5.5 \times 10^{-7}}{2 \times 10^{-4} \times \frac{11}{9}} \\ & =\frac{9 \times 0.6 \times 5.5 \times 10^{-3}}{2 \times 11} \\ & =1.35 \times 10^{-3} \mathrm{~m}=1.35 \mathrm{~mm} \end{aligned} $

Given, wavelength, $\lambda=5000 \mathop A\limits^o=5 \times 10^{-7} \mathrm{~m}$

Slit separation, $d=3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}$

Distance between slit and screen,

$D=20 \mathrm{~cm}=0.2 \mathrm{~m}$

Thickness of transparent plate $=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}$ Fringe shift, $\Delta x=6 \mathrm{~mm}=6 \times 10^{-3} \mathrm{~m}$

Fringe shift is given by

$\begin{aligned} & \Delta x=\frac{D}{d}(\mu-1) t \\ & \Rightarrow \quad 6 \times 10^{-3}=\frac{0.2}{3 \times 10^{-3}}(\mu-1) \times 10^{-3} \\ & \Rightarrow \frac{6 \times 3 \times 10^{-3}}{0.2}=\mu-1 \Rightarrow 9 \times 10^{-2}=\mu-1 \\ & \Rightarrow \quad \mu=1.09 \\ \end{aligned}$

Hence, refractive index of transparent plate is 1.09.

Given :

$\begin{aligned} \lambda_1 &= 3750 \, \mathring{A} \\ \lambda_2 &= 7500 \, \mathring{A} \\ D &= 4 \, \mathrm{m} \\ d &= 3 \, \mathrm{mm} = 3 \times 10^{-3} \, \mathrm{m} \end{aligned}$

Since,

$ \frac{\lambda_1}{\lambda_2} = \frac{3750}{7500} = \frac{1}{2} $

We know that,

$ \Rightarrow n_1 \lambda_1 = n_2 \lambda_2 \Rightarrow \frac{n_2}{n_1} = \frac{\lambda_1}{\lambda_2} = \frac{1}{2} $

Therefore, the required minimum distance is:

$\begin{aligned} x &= \frac{n_1 \lambda_1 D}{d} \\ &= \frac{2 \times 3750 \times 10^{-10} \times 4}{3 \times 10^{-3}} \\ &= \frac{3 \times 10^4 \times 10^{-10}}{3 \times 10^{-3}} \\ &= 10^{-3} \, \mathrm{m} = 1 \, \mathrm{mm} \end{aligned}$

For the given condition,

$\frac{n_1 \lambda_1}{x_1}=\frac{n_2 \lambda_2}{x_2} \quad \text{.... (i)}$

Here, $n_1=5$ (for fifth order)

$\begin{aligned} & \lambda_1=600 \mathrm{~nm}=6 \times 10^{-7} \mathrm{~m} \\ & x_1=6 \mathrm{~mm}=6 \times 10^{-3} \mathrm{~m} \\ & n_2=3(\text { for third order }) \\ & \lambda_2=400 \mathrm{~nm}=4 \times 10^{-7} \mathrm{~m} \\ & x_2=? \end{aligned}$

Putting these values in Eq. (i), we get

$\frac{5 \times 6 \times 10^{-7}}{6 \times 10^{-3}}=\frac{3 \times 4 \times 10^{-7}}{x_2}$

$\begin{aligned} & \Rightarrow \quad 5 x_2=12 \times 10^{-3} \\ & \Rightarrow \quad x_2=2.4 \times 10^{-3} \mathrm{~m} \\ & \qquad =2.4 \mathrm{~mm} \\ & \end{aligned}$

As, shift $\Delta x = {\beta \over \lambda }(\mu - 1)t$

Shift due to one plate $\Delta {x_1} = {\beta \over \lambda }({\mu _1} - 1)t$

Shift due to another path $\Delta {x_2} = {\beta \over \lambda }({\mu _2} - 1)t$

Net shift $\Delta x = \Delta {x_2} - \Delta {x_1} = {\beta \over \lambda }({\mu _2} - {\mu _1})t$ ..... (i)

Also, it is given that $\Delta$x = 6$\beta$

Hence, $6\beta = {\beta \over \lambda }({\mu _1} - {\mu _2})t$

$ \Rightarrow t = {{6\lambda } \over {({\mu _1} - {\mu _2})}} = {{6 \times 4000} \over {(1.7 - 1.4)}} - 80000\,\mathop A\limits^o $

$ \Rightarrow t = 8 \times {10^{ - 6}} = 8\mu $m

The answer is Option B : the distance between consecutive fringes will decrease.

Young's double-slit experiment depends on the principle of interference of light waves, and the resulting pattern of light and dark bands (fringes) depends on the wavelength of the light used.

The formula for the distance between fringes (y) in the double slit experiment is :

y = ${{\lambda D} \over d}$

where :

• λ is the wavelength of the light
• D is the distance from the slits to the screen
• d is the distance between the slits

The wavelength of orange light is approximately 600 nm while the wavelength of blue light is around 475 nm. Therefore, if you change the light from orange to blue, the wavelength λ decreases. As λ is directly proportional to y, if λ decreases, y will also decrease. So, the distance between consecutive fringes will decrease.

In the double-slit experiment, the position of a bright fringe is given by the formula :

$y = \frac{{m\lambda L}}{{d}}$,

where :

• m is the order of the fringe (1 for the first bright fringe, 2 for the second, etc.)
• $\lambda$ is the wavelength of the light (in meters)
• L is the distance from the slits to the screen (in meters)
• d is the distance between the slits (in meters)

We need to find the difference in position between the first and third bright fringes. So, we find the position of both and subtract the position of the first from the position of the third :

$y_3 = \frac{{3\lambda L}}{{d}}$

$y_1 = \frac{{\lambda L}}{{d}}$

Then, the difference between the third and the first bright fringes is :

$y_3 - y_1 = 2\frac{{\lambda L}}{{d}}$

Now, let's plug the given values: $\lambda = 5890 \,Å = 5890 \times 10^{-10} \, m$ (since 1 Å = $10^{-10}$ meters), L = 0.5 m, and d = 0.5 mm = 0.5 $\times 10^{-3}$ m :

$ y_3 - y_1 = 2 \times \frac{5890 \times 10^{-10} \, \text{m} \times 0.5 \, \text{m}}{0.5 \times 10^{-3} \, \text{m}} = 1178 \times 10^{-6} \, \text{m} $

So, the correct answer is 1178 $\times 10^{-6}$ m, which corresponds to Option B.

Yes, you're absolutely correct. This equation you provided is for the angular size of the central maximum (or central diffraction disk) in a circular aperture diffraction pattern (like a pinhole) :

$ \sin \theta = \frac{1.22\lambda}{D} $

where :

• θ is the angle subtended by the radius of the central maximum at the pinhole,


• λ is the wavelength of the light, and


• D is the diameter of the aperture or pinhole.

If D (the diameter of the pinhole) is increased, then sinθ (and hence θ itself, for small θ) will decrease, implying that the size of the central maximum or diffraction disk will decrease. This is because less diffraction (bending of light) occurs when the pinhole is larger.

At the same time, increasing the size of the pinhole allows more light to pass through, which increases the intensity (brightness) of the light in the diffraction pattern.

So, increasing the diameter of the pinhole decreases the size of the diffraction pattern but increases its intensity, which confirms Option C.

As we know that, wavefront is defined as the locus of all points in same phase.

As we know that,

for diffraction, $d\sin\theta=n\lambda$

where, d is slit width, $\theta$ is angular width and $\lambda$ is wavelength.

Now, if $\lambda$ decreases then $\sin\theta$ decreases and pattern shrinks. So, images moves towards O.

Initially, separation between slits $=d$

Distance of slit from screen $=D$

Initial and final fringe widths be $\beta$ and $\beta^{\prime}$.

Finally, separation between slits, $d^{\prime}=d / 2$

Distance from screen $D^{\prime}=2 D$

As,

$\begin{aligned} \beta & =\frac{\lambda D}{d} \\ \therefore \beta^{\prime} & =\frac{\lambda D^{\prime}}{d^{\prime}}=\frac{\lambda 2 D}{d / 2}=\frac{4 \lambda D}{d} \end{aligned}$

$\Rightarrow \quad \beta^{\prime}=4 \beta$

Angular width for Ist minima, $\theta_1=30^\circ=\pi/6$ rad

Wavelength, $\lambda=500$ nm = $500\times10^{-9}$ m

Let first secondary maximum occur at angle $\theta_2$.

$\because$ $n\lambda = a\sin {\theta _1}$

where, a is slit width.

$\therefore$ $a = {{n\lambda } \over {\sin 30\Upsilon }} = 2\lambda $ [$\because$ $n = 1$]

and for maxima

$(2n + 1){\lambda \over 2} = a\sin {\theta _2}$

$ \Rightarrow (2 + 1){\lambda \over 2} = a\sin {\theta _2}$

$ \Rightarrow \sin {\theta _2} = 3\lambda /2a$ ..... (i)

Substituting the value $a=2\pi$ in Eq. (i), we get

$\begin{array}{ll}\therefore & \sin \theta_2=\frac{3 \lambda}{2 \times 2 \lambda}=\frac{3}{4} \\ \Rightarrow & \theta_2=\sin ^{-1}(3 / 4)\end{array}$

As we know that, Resolving power of optical instrument,

$ \mathrm{RP}=\frac{2 \mu \sin \theta}{1.22 \lambda} \Rightarrow \mathrm{RP} \text { or } R \propto \frac{1}{\lambda} $

Hence, ratio of resolving power of two instruments is

$ \begin{aligned} & \frac{R_1}{R_2}=\frac{\lambda_2}{\lambda_1} \Rightarrow \frac{R_1}{R_2}=\frac{5000}{4000} \\ & {\left[\because \lambda_1=4000 \mathop {\rm{A}}\limits^{\rm{o}} \text { and } \lambda_2=5000 \mathop {\rm{A}}\limits^{\rm{o}}\right]} \\ & \frac{R_1}{R_2}=\frac{5}{4} \text { or } 5: 4 \end{aligned} $

Fringe width in Young's double slit experiment is given as

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

where, $D=$ distance between source and screen and $d=$ slit width.

$ \begin{array}{ll} \therefore & \frac{\beta_2}{\beta_1}=\frac{d_1}{d_2} \times \frac{D_2}{D_1} \\ \Rightarrow & \frac{\beta_2}{\beta_1}=\frac{d_1}{2 d_1} \times \frac{D_1 / 2}{D_1} \\ & \quad\left[\because d_2=2 d_1 \text { and } D_2=\frac{D_1}{2}\right] \\ \Rightarrow & \frac{\beta_2}{\beta_1}=\frac{1}{4} \Rightarrow B_2=\frac{\beta_1}{4} \end{array} $

Limit of resolution or angular resolution for a telescope is

$ \alpha_{\min }=\frac{1.22 \lambda}{D} $

Here, $\alpha_{\text {min }}=3 \times 10^{-7} \mathrm{rad}$,

$ \lambda=525 \mathrm{~nm}=525 \times 10^{-9} \mathrm{~m} $

∴ Diameter of objective lens is

$ \begin{aligned} D & =\frac{1.22 \lambda}{\alpha_{\min }} \\ & =\frac{1.22 \times 525 \times 10^{-9}}{3 \times 10^{-7}} \\ & =213.5 \times 10^{-2}=2.1 \mathrm{~m} \end{aligned} $

Given, $\lambda_1=6600 \mathop {\rm{A}}\limits^{\rm{o}}, \lambda_2=4400 \mathop {\rm{A}}\limits^{\rm{o}}$,

$ n_1=60 $

For same set up of YDSE,

$ \begin{aligned} n_1 \lambda_1 & =n_2 \lambda_2 \\ \Rightarrow n_2 & =\frac{n_1 \lambda_1}{\lambda_2}=\frac{60 \times 6600}{4400} \\ & =\frac{60 \times 6}{4}=15 \times 6=90 \text { fringes } \end{aligned} $

$ \begin{aligned} &\text { Fringe width in YDSE is given by }\\ &\begin{aligned} & \beta=\frac{\lambda D}{d} \Rightarrow \beta \propto \lambda \\ & \text { As, } \lambda_R>\lambda_G>\lambda_B \\ & \Rightarrow \beta_R>\beta_G>\beta_B \end{aligned} \end{aligned} $