Answer

Let $I_{1}$ and $I_{2}$ be the intensity of the two light waves. Their resultant intensities can be obtained as:

$I^{\prime}=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \cos \phi$

Where,

$\phi=$ Phase difference between the two waves

For monochromatic light waves,

$$ \begin{aligned} & I_{1}=I_{2} \\ & \begin{aligned} \therefore I^{\prime} & =I_{1}+I_{1}+2 \sqrt{I_{1} I_{1}} \cos \phi \\ & =2 I_{1}+2 I_{1} \cos \phi \end{aligned} \end{aligned} $$

Phase difference $=\frac{2 \pi}{\lambda} \times$ Path difference

Since path difference $=\lambda$,

Phase difference, $\phi=2 \pi$

$\therefore I^{\prime}=2 I_{1}+2 I_{1}=4 I_{1}$

Given,

$I^{\prime}=K$

$\therefore I_{1}=\frac{K}{4}$

When path difference $=\frac{\lambda}{3}$,

Phase difference, $\phi=\frac{2 \pi}{3}$

Hence, resultant intensity, $I_{R}^{\prime}=I_{1}+I_{1}+2 \sqrt{I_{1} I_{1}} \cos \frac{2 \pi}{3}$

$=2 I_{1}+2 I_{1}\left(-\frac{1}{2}\right)=I_{1}$

Using equation (1), we can write:

$I_{R}=I_{1}=\frac{K}{4}$

Hence, the intensity of light at a point where the path difference is $\frac{\lambda}{3}$ is $\frac{K}{4}$ units.

Answer

The shape of the wavefront in case of a light diverging from a point source is spherical. The wavefront emanating from a point source is shown in the given figure.

The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid. This is shown in the given figure.

The portion of the wavefront of light from a distant star intercepted by the Earth is a plane.

Answer Refractive index of glass, $\mu=1.5$

Speed of light, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

Speed of light in glass is given by the relation,

$$ \begin{aligned} v & =\frac{c}{\mu} \\ & =\frac{3 \times 10^{8}}{1.5}=2 \times 10^{8} \mathrm{~m} / \mathrm{s} \end{aligned} $$

Hence, the speed of light in glass is $2 \times 10^{8} \mathrm{~m} / \mathrm{s}$.

The speed of light in glass is not independent of the colour of light.

The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.

Answer

Distance between the slits, $d=0.28 \mathrm{~mm}=0.28 \times 10^{-3} \mathrm{~m}$

Distance between the slits and the screen, $D=1.4 \mathrm{~m}$

Distance between the central fringe and the fourth $(n=4)$ fringe,

$u=1.2 \mathrm{~cm}=1.2 \times 10^{-2} \mathrm{~m}$

In case of a constructive interference, we have the relation for the distance between the two fringes as:

$u=n \lambda \frac{D}{d}$

Where,

$n=$ Order of fringes $=4$ $\lambda=$ Wavelength of light used

$$ \therefore=\frac{u d}{n D} $$

$=\frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4}$

$=6 \times 10^{-7}$

$=600 \mathrm{~nm}$

Hence, the wavelength of the light is $600 \mathrm{~nm}$.

Answer

Let $I_{1}$ and $I_{2}$ be the intensity of the two light waves. Their resultant intensities can be obtained as:

$I^{\prime}=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \cos \phi$

Where,

$\phi=$ Phase difference between the two waves

For monochromatic light waves,

$$ \begin{aligned} & I_{1}=I_{2} \\ & \begin{aligned} \therefore I^{\prime} & =I_{1}+I_{1}+2 \sqrt{I_{1} I_{1}} \cos \phi \\ & =2 I_{1}+2 I_{1} \cos \phi \end{aligned} \end{aligned} $$

Phase difference $=\frac{2 \pi}{\lambda} \times$ Path difference

Since path difference $=\lambda$,

Phase difference, $\phi=2 \pi$

$\therefore I^{\prime}=2 I_{1}+2 I_{1}=4 I_{1}$

Given,

$I^{\prime}=K$

$\therefore I_{1}=\frac{K}{4}$

When path difference $=\frac{\lambda}{3}$,

Phase difference, $\phi=\frac{2 \pi}{3}$

Hence, resultant intensity, $I_{R}^{\prime}=I_{1}+I_{1}+2 \sqrt{I_{1} I_{1}} \cos \frac{2 \pi}{3}$

$=2 I_{1}+2 I_{1}\left(-\frac{1}{2}\right)=I_{1}$

Using equation (1), we can write:

$I_{R}=I_{1}=\frac{K}{4}$

Hence, the intensity of light at a point where the path difference is $\frac{\lambda}{3}$ is $\frac{K}{4}$ units.

Answer

Wavelength of the light beam, $\lambda_{1}=650 \mathrm{~nm}$

Wavelength of another light beam, $\lambda_{2}=520 \mathrm{~nm}$

Distance of the slits from the screen $=D$

Distance between the two slits $=d$

Distance of the $n^{\text {th }}$ bright fringe on the screen from the central maximum is given by the relation,

$x=n \lambda_{1}\left(\frac{D}{d}\right)$

For third bright fringe, $n=3$

$\therefore x=3 \times 650 \frac{D}{d}=1950\left(\frac{D}{d}\right) \mathrm{nm}$

Let the $n^{\text {th }}$ bright fringe due to wavelength $\lambda_{2}$ and $(n-1)^{\text {th }}$ bright fringe due to wavelength $\lambda_{1}$ coincide on the screen. We can equate the conditions for bright fringes as:

$$ \begin{aligned} & n \lambda_{2}=(n-1) \lambda_{1} \\ & 520 n=650 n-650 \\ & 650=130 n \\ & \therefore n=5 \end{aligned} $$

Hence, the least distance from the central maximum can be obtained by the relation:

$$ \begin{aligned} x & =n \lambda_{2} \frac{D}{d} \\ & =5 \times 520 \frac{D}{d}=2600 \frac{D}{d} \mathrm{~nm} \end{aligned} $$

Note: The value of $d$ and $D$ are not given in the question.