5 Marks Questions

Question 15 Marks

A mica strip and a polysterene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50mm and the separation between the slits is 0.12cm. The refractive index of mica and polysterene are 1.58 and 1.55 respectively for the light of wavelength 590nm which is used in the experiment. The interference is observed on a screen a distance one meter away.

What would be the fringe-width?
At what distance from the centre will the first maximum be located?

Answer

Given that, t₁ = t₂ = 0.5mm = 0.5 × 10^-3m, $\mu_\text{m}=1.58$ and $\mu_\text{p}=1.55,$

$\lambda=590\text{nm}=590\times 10^{-9}\text{m}, \text{d}=0.12\text{cm},\\ \mu_{\text{m}}=1.58\text{ and }\mu_{\text{p}}=1.55,$

Fringe width $=\frac{\text{D}\lambda}{\text{d}}=\frac{1\times590\times10^{-9}}{12\times10^{-4}}=4.91\times10^{-4}\text{m}.$

When both the strips are fitted, the optical path changes by

$\Delta\text{x}=(\mu_\text{m}-1)\text{t}_1-(\mu_\text{p}-1)\text{t}_2=(\mu_\text{m}-\mu_\text{p})\text{t}$

$=(1.58-1.55)\times(0.5)(10^{-3})=0.015\times10^{-13}\text{m}$

So, No. of fringes shifted $=\frac{0.015\times10^{-3}}{590\times10^{-3}}=25.43$

⇒ There are 25 fringes and 0.43 th of a fringe.

⇒ There are 13 bright fringes and 12 dark fringes and 0.43 th of a dark fringe.

So, position of first maximum on both sides will be given by,

$\therefore$ x = 0.43 × 4.91 × 10^-4 = 0.021cm

x' = (1 - 0.43) × 4.91 × 10^-4 = 0.028cm (since, fringe width = 4.91 × 10^-4m)

View full question & answer→

Question 25 Marks

Consider the situation shown in figure (17-E6). The two slits S₁ and S₂ p laced symmetrically around the central line are illuminated by a monochromatic light of wavelength $\lambda.$ The separation between the slits is d. The light transmitted by the slits falls on a screen E₁ placed at a distance D from the slits. The slit S₃ is at the central line and the slit S₄ is at a distance z from S₃. Another

screen $\sum_2$ is placed a further distance D away from $\sum_1$. Find the ratio of the maximum to minimum intensity observed on $\sum_2$, if z is equal to,

$\text{z}=\frac{\lambda\text{D}}{2\text{d}}$
$\frac{\lambda\text{D}}{\text{d}}$
$\frac{\lambda\text{D}}{4\text{d}}$

Answer

When, $\text{z}=\frac{\lambda\text{D}}{2\text{d}},$ at S₄, minimum intensity occurs

⇒ Amplitude = 0,

At S₃, path difference = 0

⇒ Maximum intensity occurs.

⇒ Amplitude = 2r.

So, on $\sum_2$ screen,

$\frac{\text{I}_\text{max}}{\text{I}_\text{min}}=\frac{(2\text{r}+0)^2}{(2\text{r}-0)^2}=1$

When, $\text{z}=\frac{\lambda\text{D}}{2\text{d}},$ At S₄, minimum intensity occurs.

⇒ Amplitude = 0.

At S₃, path difference = 0

⇒ Maximum intensity occurs.

⇒ Amplitude = 2r.

So, on $\sum_2$ screen,

$\frac{\text{I}_\text{max}}{\text{I}_\text{min}}=\frac{(2\text{r}+2\text{r})^2}{(2\text{r}-0)^2}=\infty$

When, $\text{z}=\frac{\lambda\text{D}}{2\text{d}},$, At S₄, intensity $=\frac{\text{I}_\text{max}}{2}$

⇒ Amplitude $=\sqrt{2\text{r}}$

$\therefore$ At S₃, intensity is maximum

⇒ Amplitude = 2r

$\therefore\frac{\text{I}_\text{max}}{\text{I}_\text{min}}=\frac{(2\text{r}+\sqrt{2\text{r}})^2}{(2\text{r}-\sqrt{2\text{r}})^2}=34.$

View full question & answer→

Question 35 Marks

In a Young's double slit interference experiment the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength $\lambda$. Find the distance from the central point where the intensity falls to,

Half the maximum.
One fourth of the maximum.

Answer

When intensity is half the maximum $\frac{\text{I}}{\text{I}_\text{max}}=\frac{1}{2}$

$\Rightarrow\frac{4\text{a}^2\cos^2\big(\frac{\phi}{2}\big)}{4\text{a}^2}=\frac{1}{2}$

$\Rightarrow\cos^2\Big(\frac{\phi}{2}\Big)=\frac{1}{2}\Rightarrow\cos\Big(\frac{\phi}{2}\Big)=\frac{1}{\sqrt{2}}$

$\Rightarrow\frac{\phi}{2}=\frac{\pi}{4}\Rightarrow\phi=\frac{\pi}{2}$

$\Rightarrow$ Path difference, $\text{x}=\frac{\lambda}{4}$

$\Rightarrow\text{y}=\frac{\text{xD}}{\text{d}}=\frac{\lambda\text{D}}{4\text{d}}$

When intensity is $\frac{1}{4}\text{th}$ of the maximum $\frac{\text{I}}{\text{I}_\text{max}}=\frac{1}{4}$

$\Rightarrow\frac{4\text{a}^2\cos^2\Big(\frac{\phi}{2}\Big)}{4\text{a}^2}=\frac{1}{4}$

$\Rightarrow\cos^2\Big(\frac{\phi}{2}\Big)=\frac{1}{4}\Rightarrow\cos\Big(\frac{\phi}{2}\Big)=\frac{1}{2}$

$\Rightarrow\frac{\phi}{2}=\frac{\pi}{3}\Rightarrow\phi=\frac{2\pi}{3}$

$\Rightarrow$ Path difference, $\text{x}=\frac{\lambda}{3}$

$\Rightarrow\text{y}=\frac{\text{xD}}{\text{d}}=\frac{\lambda\text{D}}{3\text{d}}$

View full question & answer→

Question 45 Marks

Consider the arrangement shown in figure. By some mechanism, the separation between the slits S₃ and S₄ can be changed. The intensity is measured at the point P which is at the common perpendicular bisector

of S₁S₂ and S₃S₄. When $\text{z}=\frac{\text{D}\lambda}{2\text{d}},$ intensity measured at P is I. Find this intensity when z is equal to:

$\frac{\text{D}\lambda}{\text{d}}$
$\frac{3\text{D}\lambda}{2\text{d}}$
$\frac{2\text{D}\lambda}{\text{d}}$

Answer

When, $\text{z}=\frac{\text{D}\lambda}{\text{d}}$

So, $\text{OS}_3=\text{OS}_4=\frac{\text{D}\lambda}{2\text{d}}$

⇒ Dark fringe at S₃ and S₄.

⇒ At S₃, intensity at S₃ = 0 ⇒ I₁ = 0

At S₄, intensity at S₄ = 0 ⇒ I₂ = 0

At P, path difference = 0 ⇒ Phase difference = 0.

$\Rightarrow\text{I}=\text{I}_1+\text{I}_2+\sqrt{\text{I}_1\text{I}_2}\cos0^\circ=0+0+0=0$

⇒ Intensity at P = 0.

Given that, when $\text{z}=\frac{\text{D}\lambda}{2\text{d}}$, intensity at P = I

Here, $\text{OS}_3=\text{OS}_4=\text{y}=\frac{\text{D}\lambda}{4\text{d}}$

$\therefore\phi=\frac{2\pi\text{x}}{\lambda}=\frac{2\pi}{\lambda}\times\frac{\text{yd}}{\text{D}}=\frac{2\pi}{\lambda}\times\frac{\text{D}\lambda}{4\text{d}}\times\frac{\text{d}}{\text{D}}=\frac{\pi}{2}.$

$\Big[$Since, x = path difference $=\frac{\text{yd}}{\text{D}}\Big]$

Let, intensity at S₃ and S₄ = I'

$\therefore$ At P, phase difference = 0

So, $\text{I}'+\text{I}'+2\text{I}'\cos0^\circ=\text{I}$

$\Rightarrow4\text{I}'=\text{I}\Rightarrow\text{I}'=\frac{1}{4}$

When, $\text{z}=\frac{3\text{D}\lambda}{2\text{d}},\Rightarrow\text{y}=\frac{3\text{D}\lambda}{4\text{d}}$

$\therefore\phi=\frac{2\pi\text{x}}{\lambda}=\frac{2\pi}{\lambda}\times\frac{\text{yd}}{\text{D}}=\frac{2\pi}{\lambda}\times\frac{3\text{D}\lambda}{4\text{d}}\times\frac{\text{d}}{\text{D}}=\frac{3\pi}{2}$

Let, I'' be the intensity at S₃ and S₄ when, $\phi=\frac{3\pi}{2}$

Now comparing,

$\frac{\text{I}"}{\text{I}}=\frac{\text{a}^2+\text{a}^2+2\text{a}^2\cos\Big(\frac{3\pi}{2}\Big)}{\text{a}^2+\text{a}^2+2\text{a}^2\cos\frac{\pi}{2}}=\frac{2\text{a}^2}{2\text{a}^2}=1$ $\Rightarrow\text{I}"=\text{I}'=\frac{\text{I}}{4}$

$\therefore$ Intensity at $\text{P}=\frac{\text{I}}{4}+\frac{\text{I}}{4}+2\times\Big(\frac{\text{I}}{4}\Big)\cos0^\circ=\frac{\text{I}}{2}+\frac{\text{I}}{2}=1$

When $\text{z}=\frac{2\text{D}\lambda}{\text{d}}$

$\Rightarrow\text{y}=\text{OS}_3=\text{OS}_4=\frac{\text{D}\lambda}{\text{d} }$

$\therefore\phi=\frac{2\pi\text{X}}{\lambda}=\frac{2\pi}{\lambda}\times\frac{\text{yd}}{\text{D}}=\frac{2\pi}{\lambda}\times\frac{\text{D}\lambda}{\text{d}}\times\frac{\text{d}}{\text{D}}=2\pi.$

Let, I"' = intensity at S₃ and S₄ when, $\phi=2\pi.$

$\frac{\text{I}'''}{\text{I}'}=\frac{\text{a}^2+\text{a}^2+2\text{a}^2\cos2\pi}{\text{a}^2+\text{a}^2+2\text{a}^2\cos\frac{\pi}{2}}=\frac{4\text{a}^2}{2\text{a}^2}=2$

$\Rightarrow\text{I}'''=2\text{I}'=2\Big(\frac{\text{I}}{4}\Big)=\frac{\text{I}}{2}$

At P, I_resultant $=\frac{\text{I}}{2}+\frac{\text{I}}{2}+2\Big(\frac{\text{I}}{2}\Big)\cos0^\circ=\text{I}+\text{I}=2\text{I}$

So, the resultant intensity at P will be 2I.

View full question & answer→

Question 55 Marks

A thin paper of thickness 0.02mm having a refractive index 1.45 is pasted across one of the slits in a Young's double slit experiment. The paper transmits $\frac{4}{9}$ of the light energy falling on it.

Find the ratio of the maximum intensity to the minimum intensity in the fringe pattern.
How many fringes will cross through the centre if an identical paper piece is pasted on the other slit also? The wavelength of the light used is 600nm.

Answer

Given that, t = 0.02mm = 0.02 × 10^-3m, $\mu_1=1.45,\ \lambda=600\text{nm}=600\times10^{-9}\text{m}$

Let, I₁ = Intensity of source without paper = I
Then I₂ = Intensity of source with paper $=\Big(\frac{4}{9}\Big)\text{I}$

$\Rightarrow\frac{\text{I}_1}{\text{I}_2}=\frac{9}{4}\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{3}{2}\ [\because\text{I}\propto\text{r}^2]$

where, r₁ and r₂ are corresponding amplitudes.

So, $\frac{\text{I}_\text{max}}{\text{I}_\text{min}}=\frac{(\text{r}_1+\text{r}_2)^2}{(\text{r}_1-\text{r}_2)^2}=25:1$

No. of fringes that will cross the origin is given by,

$\text{n}=\frac{(\mu-1)\text{t}}{\lambda}=\frac{(1.45-1)\times0.02\times10^{-3}}{600\times10^{-9}}=15.$

View full question & answer→

Question 65 Marks

Figure shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let $\text{BP}_0-\text{AP}=\frac{\lambda}{3}$ and $\text{D}>\lambda.$

Show that in this case $\text{d}=\sqrt{\frac{2\lambda\text{D}}{3}}.$
Show that the intensity at P, is three times the intensity due to any of the three slits individually.

Answer

Since S₁, S₂ are in same phase, at O there will be maximum intensity.

Given that, there will be a maximum intensity at P.

⇒ path difference $=\Delta\text{x}=\text{n}\lambda$

From the figure,

$(\text{S}_1\text{P})^2-(\text{S}_2\text{P})^2=\Big(\sqrt{\text{D}^2+\text{X}^2}\Big)^2-\Big(\sqrt{(\text{D}-2\lambda)^2+\text{X}^2}\Big)^2$

$=4\lambda\text{D}-4\lambda^2=4\lambda\text{D}$ ($\lambda^2$ is so small and can be neglected)

$\Rightarrow\text{S}_1\text{P}-\text{S}_2\text{P}=\frac{4\lambda\text{D}}{2\sqrt{\text{x}^2+\text{D}^2}}=\text{n}\lambda$

$\Rightarrow\frac{2\text{D}}{\sqrt{\text{x}^2+\text{D}^2}}=\text{v}$

$\Rightarrow\text{n}^2(\text{X}^2+\text{D}^2)=4\text{D}^2=\Delta\text{X}=\frac{\text{D}}{\text{n}}\sqrt{4-\text{n}^2}$

when $\text{n}=1,\text{x}=\sqrt{3}\text{D}$ (1^st order)

$\text{n}=2,\text{x}=0$ (2^nd order)

$\therefore$ When $\text{X}=\sqrt{3}\text{D},$ at P there will be maximum intensity.

View full question & answer→

Question 75 Marks

Two coherent point sources S₁ and S₂ vibrating in phase emit light of wavelength $\lambda$. The separation between the sources is $2\lambda$. Consider a line passing through S₂ and perpendicular to the line S₁S₂. What is the smallest distance from S₂ where a minimum of intensity occurs?

Answer

For minimum intensity

$\therefore\text{S}_1\text{P}-\text{S}_2\text{P}=\text{x}=(2\text{n}+1)\frac{\lambda}{2}$

From the figure, we get,

$\Rightarrow\sqrt{\text{X}^2+(2\lambda)^2}-\text{Z}=(2\text{n}+1)\frac{\lambda}{2}$

$\Rightarrow\text{Z}^2+4\lambda^2=\text{Z}^2+(2\text{n}+1)^2\frac{\lambda^2}{4}+\text{Z}(2\text{n}+1)\lambda$

$\Rightarrow\text{Z}=\frac{4\lambda^2-(2\text{n}+1)^2\Big(\lambda^2{4}\Big)}{(2\text{n}+1)\lambda}=\frac{16\lambda^2-(2\text{n}+1)^2\lambda^2}{4(2\text{n}+1)\lambda}\ ...(1)$

Putting, $\text{n}=0\Rightarrow\text{Z}=\frac{15\lambda}{4}$

$\text{n}=-1\Rightarrow\text{Z}=\frac{-15\lambda}{4}$

$\text{n}=1\Rightarrow\text{Z}=\frac{7\lambda}{12}$

$\text{n}=2\Rightarrow\text{Z}=\frac{-9\lambda}{20}$

$\therefore\text{Z}=\frac{7\lambda}{12}$ is the smallest distance for which there will be minimum intensity.

View full question & answer→

Question 85 Marks

White coherent light (400nm - 700nm) is sent through the slits of a Young's double slit experiment (figure.) The separation between the slits is 0.5mm and the screen is 50cm away from the slits. There is a hole in the screen at a point 1.0mm away (along the width of the fringes) from the central line.

Which wavelength (s) will be absent in the light coming from the hole?
which wavelength (s) will have a strong intensity?

Answer

Given that, $\lambda= (400\text{nm to } 700\text{nm}),$ d = 0.5mm = 0.5 × 10^-3m,

D = 50cm = 0.5m and on the screen y_n = 1mm = 1 × 10^-3m

We know that for zero intensity (dark fringe)

$\text{y}_\text{n}=\Big(\frac{2\text{n}+1}{2}\Big)\frac{\lambda_\text{n}\text{D}}{\text{d}}$ where n = 0, 1, 2, ...

$\Rightarrow\lambda_\text{n}=\frac{2}{(2\text{n}+1)}\frac{\lambda_\text{n}\text{d}}{\text{D}}=\frac{2}{2\text{n}+1}\times\frac{10^{-3}\times0.5\times10^{-3}}{0.5}$

$\Rightarrow\frac{2}{(2\text{n}+1)}\times10^{-6}\text{m}=\frac{2}{(2\text{n}+1)}\times10^3\text{nm}$

If $\text{n}=1,\lambda_1=\Big(\frac{2}{3}\Big)\times1000=667\text{nm}$

If $\text{n}=1,\lambda_2=\Big(\frac{2}{5}\Big)\times1000=400\text{nm}$

So, the light waves of wavelengths 400nm and 667nm will be absent from the out coming light.

For strong intensity (bright fringes) at the hole

$\text{y}_\text{n}=\frac{\text{n}\lambda_\text{n}\text{D}}{\text{d}}\Rightarrow\lambda_\text{n}=\frac{\text{y}_\text{n}\text{d}}{\text{nD}}$

When, $\text{n}=1,\lambda_1=\frac{\text{y}_\text{n}\text{d}}{\text{D}}$

$=\frac{10^{-3}\times0.5\times10^{-3}}{0.5}=10^{-6}\text{m}=1000\text{nm}.$

1000nm is not present in the range 400nm - 700nm

Again, where $\text{n}=2,\lambda_2=\frac{\text{y}_\text{n}\text{d}}{2\text{D}}=500\text{nm}$

So, the only wavelength which will have strong intensity is 500nm.

View full question & answer→

Question 95 Marks

A glass surface is coated by an oil film of uniform thickness 1.00 × 10^-4cm. The index of refraction of the oil is 1.25 and that of the glass is 1.50. Find the wavelengths of light in the visible region (400nm - 750nm) which are completely transmitted by the oil film under normal incidence.

Answer

For the thin oil film,

$\text{d}=1\times10^{-4}\text{cm}=10^{-6}\text{m},$ $\mu_{\text{oil}}=1.25\ \text{and}\ \mu_\text{x}=1.50$

$\lambda=\frac{2\mu\text{d}}{\Big(\text{n}+\frac{1}{2}\Big)}\frac{2\times10^{-6}\times1.25\times2}{2\text{n}+1}=\frac{5\times10^{-6}\text{m}}{2\text{n}+1}$

$\Rightarrow\lambda=\frac{5000\text{nm}}{2\text{n}+1}$

For the wavelengths in the region (400nm - 750nm)

When, $\text{n}=3,\lambda=\frac{5000}{2\times3+1}=\frac{5000}{7}=714.3\text{nm}$

When, $\text{n}=4,\lambda=\frac{5000}{2\times4+1}=\frac{5000}{9}=555.6\text{nm}$

When, $\text{n}=5,\lambda=\frac{5000}{2\times5+1}=\frac{5000}{11}=454.5\text{nm}$

View full question & answer→

Question 105 Marks

A parallel beam of white light is incident normally on a water film 1.0 × 10^-4cm thick. Find the wavelength in the visible range (400nm - 700nm) which are strongly transmitted by the film. Refractive index of water = 1.33.

Answer

For strong transmission, $2\mu\text{d}=\text{n}\lambda\Rightarrow\lambda=\frac{2\mu\text{d}}{\text{n}}$

Given that, $\mu=1.33,\text{d}=1\times10^{-4}\text{cm}=1\times10^{-6}\text{m}$

$\Rightarrow\lambda=\frac{2\times1.33\times1\times10^{-6}}{\text{n}}=\frac{2660\times10^{-9}}{\text{n}}\text{m}$

When, $\text{n}=4,\lambda_1=665\text{nm}$

$\text{n}=5,\lambda_2=532\text{nm}$

$\text{n}=6,\lambda_3=443\text{nm}$

View full question & answer→

Question 115 Marks

Consider the arrangement shown in figure (17-E4). The distance D is large compared to the separation d between the slits.

Find the minimum value of d so that there is a dark fringe at 0.
Suppose d has this value. Find the distance x at which the next bright fringe is formed.
Find the fringe-width.

Answer

From the diagram, it can be seen that at point O.

Path difference = (AB + BO) - (AC + CO)

= 2(AB - AC) [Since, AB = BO and AC = CO] $=2\Big(\sqrt{\text{d}^2+\text{D}^2}-\text{D}\Big)$

For dark fringe, path difference should be odd multiple of $\frac{\lambda}{2}.$

So, $2\Big(\sqrt{\text{d}^2+\text{D}^2}-\text{D}\Big)=(2\text{n}+1)\Big(\frac{\lambda}{2}\Big)$

$\Rightarrow\sqrt{\text{d}^2+\text{D}^2}=\text{D}+(2\text{n}+1)\Big(\frac{\lambda}{4}\Big)$

$\Rightarrow\text{D}^2+\text{d}^2=\text{D}^2+(2\text{n}+1)^2\frac{\lambda^2}{16}+(2\text{n}+1)\frac{\lambda\text{D}}{2}$

Neglecting, $(2\text{n}+1)^2\frac{\lambda^2}{16},$ as it is very small

We get, $\text{d}=\sqrt{(2\text{n}+1)\frac{\lambda\text{D}}{2}}$

For minimum ‘d’, putting $\text{n}=0\Rightarrow\text{d}_\text{min}=\sqrt{\frac{\lambda\text{D}}{2}}.$

View full question & answer→

Question 125 Marks

In a Young's double slit experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 micron (1 micron = 10^-6m) is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the screen and the slits is doubled. It is found that the distance between the successive maxima now is the same as the observed fringe-shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment.

Answer

In the given Young’s double slit experiment,

$\mu=1.6,$ t = 1.964 micron = 1.964 × 10^-6m

We know, number of fringes shifted $=\frac{(\mu-1)\text{t}}{\lambda}$

So, the corresponding shift = No.of fringes shifted × fringe width

$=\frac{(\mu-1)\text{t}}{\lambda}\times\frac{\lambda\text{D}}{\text{d}}=\frac{(\mu-1)\text{tD}}{\text{d}}\ ...(1)$

Again, when the distance between the screen and the slits is doubled,

Fringe width $=\frac{\lambda(2\text{D})}{\text{d}}\ ...(2)$

From (1) and (2), $\frac{(\mu-1)\text{tD}}{\text{d}}=\frac{\lambda(2\text{D})}{\text{d}}$

$\Rightarrow\lambda=\frac{(\mu-1)\text{t}}{\lambda}$

$=\frac{(1.6-1)\times(1.964)\times10^{-6}}{2}=589.2\times10^{-9}=589.2\text{nm}.$

View full question & answer→

Question 135 Marks

In a Young's double slit experiment, the separation between the slits = 2.0mm, the wavelength of the light = 600nm and the distance of the screen from the slits = 2.0m. If the intensity at the centre of the central maximum is 0.20W/ m², what will be the intensity at a point 0.5cm away from this centre along the width of the fringes?

Answer

Given that, d = 2mm = 2 × 10^-3m, $\lambda=600\text{nm}=6\times10^{-7}\text{m},$ I_max = 0.20W/ m², D = 2m

For the point, y = 0.5cm

We know, path difference $=\text{x}=\frac{\text{yd}}{\text{D}}=\frac{0.5\times10^{-2}\times2\times10^{-3}}{2}=5\times10^{-6}\text{m}$

So, the corresponding phase difference is,

$\phi=\frac{2\pi\text{x}}{\lambda}=\frac{2\pi\times5\times10^{-6}}{6\times10^{-7}}\Rightarrow\frac{50\pi}{3}=16\pi+\frac{2\pi}{3}\Rightarrow\phi=\frac{2\pi}{3}$

So, the amplitude of the resulting wave at the point y = 0.5cm is,

$\text{A}=\sqrt{\text{r}^2+\text{r}^2+2\text{r}^2\cos(\frac{2\pi}{3})}=\sqrt{\text{r}^2+\text{r}^2-\text{r}^2}=\text{r}$

$\frac{\text{I}}{\text{I}_\text{max}}=\frac{\text{A}^2}{(2\text{r})^2}$ [since, maximum amplitude = 2r]

$\Rightarrow\frac{\text{I}}{0.2}=\frac{\text{A}^2}{4\text{r}^2}=\frac{\text{r}^2}{4\text{r}^2}$

$\Rightarrow\text{I}=\frac{0.2}{4}=0.05\text{W/ m}^2.$

View full question & answer→

Question 145 Marks

The wavelength of sodium light in air is 589nm.

Find its frequency in air.
Find its wavelength in water (refractive index = 1.33).
Find its frequency in water.
Find its speed in water.

Answer

Given that, for sodium light, $\lambda=589\text{nm}=589\times10^{-9}\text{m}$

$\text{f}_\text{a}=\frac{3\times10^8}{589\times10^{-9}}=5.09\times10^{14}\sec\Big[\because\text{f}=\frac{\text{c}}{\lambda}\Big]$
$\frac{\mu_\text{a}}{\mu_\text{w}}=\frac{\lambda_\text{w}}{\lambda_\text{a}}\Rightarrow\frac{1}{1.33}=\frac{\lambda_\text{w}}{589\times10^{-9}}\Rightarrow\lambda _\text{w}=443\text{nm}$
$\text{f}_\text{w}=\text{f}_\text{a}=5.09\times10^{14}\sec^{-1}$ [Frequency does not change]
$\frac{\mu_\text{a}}{\mu_\text{w}}=\frac{\text{v}_\text{w}}{\text{v}_\text{a}}\Rightarrow\text{v}_\text{w}=\frac{\mu_\text{a}\text{v}_\text{a}}{\mu_\text{w}}=\frac{3\times10^{10^8}}{1.33}=2.25\times10^8\text{m/sec}.$

View full question & answer→

Physics 5 Marks Questions

Take a timed test