NCERT Class 12 Physics — Wave Optics, Chapter 10 Exercises. All 6 questions solved.

The six exercises at the end of Chapter 10 test two things. Questions 10.1 to 10.3 are about light passing from one medium into another, and about wavefronts, the surfaces of constant phase that Huygens’ construction is built on. Questions 10.4 to 10.6 are about Young’s double-slit experiment: where the bright fringes fall on the screen, and how bright the light is between them. The arithmetic runs on three results:

• Speed in a medium — $v = c/n$. The frequency is fixed by the source and does not change on reflection or refraction, so the wavelength in the medium is $\lambda = v/\nu = \lambda_0/n$.
• Fringe positions — the $n$th bright fringe is at $x_n = n\,\dfrac{\lambda D}{d}$ from the central maximum, where $d$ is the slit separation and $D$ the distance to the screen; the fringe width is $\beta = \lambda D/d$.
• Intensity — $I = 4I_0\cos^2(\phi/2)$, where the phase difference is $\phi = \dfrac{2\pi}{\lambda}\times(\text{path difference})$.

Every length is converted to metres before it goes into a formula.

Key insight. Frequency belongs to the source, and fringes belong to path difference. Hold on to the first and Questions 10.1 and 10.3 take one line each: when light enters water or glass only its speed and wavelength change. Hold on to the second and every Young’s experiment question comes down to asking what the path difference is at a point, as a fraction of a wavelength. A whole number of wavelengths gives a bright fringe (10.4 and 10.6); a third of a wavelength gives a phase difference of $2\pi/3$ and a quarter of the peak intensity (10.5).

Question 10.1

Monochromatic light of wavelength $589\ \text{nm}$ is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is $1.33$.

Question 10.1 (a)

Solution. The reflected light travels back into the air it came from, so its speed is still $c$. Its frequency is the frequency of the source: the surface re-radiates at the frequency with which the incoming wave drives it, so reflection cannot change it. With the speed and the frequency both unchanged, the wavelength $\lambda = c/\nu$ is unchanged too. The frequency itself is

$$\nu = \frac{c}{\lambda} = \frac{3.00\times10^8}{589\times10^{-9}} = 5.09\times10^{14}\ \text{Hz}$$

Question 10.1 (b)

Solution. The frequency is again the source’s frequency, $5.09\times10^{14}\ \text{Hz}$: the number of wave crests arriving at the surface each second must equal the number leaving it, or waves would pile up at the boundary. What changes is the speed, which drops by the refractive index:

$$v = \frac{c}{n} = \frac{3.00\times10^8}{1.33} = 2.26\times10^8\ \text{m s}^{-1}$$

Since the frequency is fixed and the speed has fallen, the wavelength falls in the same ratio:

$$\lambda = \frac{v}{\nu} = \frac{\lambda_0}{n} = \frac{589\ \text{nm}}{1.33} = 443\ \text{nm}$$

Note on the book’s answer. The key prints $444\ \text{nm}$, which is what comes out of dividing the rounded speed $2.26\times10^8$ by the rounded frequency $5.09\times10^{14}$. Dividing $589$ by $1.33$ directly gives $442.9\ \text{nm}$. The difference is rounding, not an error.

Question 10.2

What is the shape of the wavefront in each of the following cases: (a) Light diverging from a point source. (b) Light emerging out of a convex lens when a point source is placed at its focus. (c) The portion of the wavefront of light from a distant star intercepted by the Earth.

Question 10.2 (a)

Solution. A wavefront joins points that the disturbance reaches at the same moment, so they are all in the same phase. Light from a point source spreads out at the same speed in every direction, so after a time $t$ it has reached every point at a distance $vt$ from the source, and those points make up a sphere centred on it.

Question 10.2 (b)

Solution. A convex lens bends every ray that comes from a point source at its focus so that it leaves parallel to the principal axis. Wavefronts are always perpendicular to the rays, and the only surface perpendicular to a bundle of parallel rays is a plane. So the spherical wavefront entering the lens leaves it as a plane wavefront.

Question 10.2 (c)

Solution. The star is a point source, so its wavefronts are spheres centred on it. By the time they reach us their radius is the distance to the star, many light-years, and the Earth intercepts only a tiny patch of such a sphere. A small patch of a very large sphere curves so little that it is indistinguishable from a flat plane, just as the ground around us looks flat although the Earth is round.

Question 10.3

(a) The refractive index of glass is $1.5$. What is the speed of light in glass? (Speed of light in vacuum is $3.0\times10^8\ \text{m s}^{-1}$.) (b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?

Question 10.3 (a)

Solution. The refractive index is defined as the ratio of the speed of light in vacuum to its speed in the medium, $n = c/v$, so

$$v = \frac{c}{n} = \frac{3.0\times10^8}{1.5} = 2.0\times10^8\ \text{m s}^{-1}$$

Question 10.3 (b)

Solution. No. The refractive index of glass depends on the wavelength of the light, and since $v = c/n$, so does the speed. (When no colour is specified, a quoted refractive index such as $1.5$ is taken to refer to yellow light.) A prism spreads white light into a spectrum because it deviates violet light more than red, and a larger deviation means a larger refractive index: $n_{\text{violet}} > n_{\text{red}}$. A larger $n$ means a smaller speed, so violet light travels more slowly in the glass than red light.

Question 10.4

In a Young’s double-slit experiment, the slits are separated by $0.28\ \text{mm}$ and the screen is placed $1.4\ \text{m}$ away. The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2\ \text{cm}$. Determine the wavelength of light used in the experiment.

Solution. At a point on the screen a distance $x$ from the centre, the path difference between the waves from the two slits is $xd/D$. A bright fringe appears wherever that path difference is a whole number of wavelengths, $n\lambda$, so the $n$th bright fringe lies at

$$x_n = n\,\frac{\lambda D}{d}$$

The central bright fringe is $n = 0$, which makes the fourth bright fringe $n = 4$, and it is at $x_4 = 1.2\ \text{cm}$. Turning the formula round, with $d = 0.28\times10^{-3}\ \text{m}$ and $D = 1.4\ \text{m}$:

$$\lambda = \frac{x_4\,d}{4D} = \frac{(1.2\times10^{-2})(0.28\times10^{-3})}{4\times1.4} = \frac{3.36\times10^{-6}}{5.6} = 6.0\times10^{-7}\ \text{m}$$

Question 10.5

In Young’s double-slit experiment using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is $\lambda$, is $K$ units. What is the intensity of light at a point where path difference is $\lambda/3$?

Solution. When two waves each of intensity $I_0$ meet with a phase difference $\phi$, the resulting intensity is

$$I = 4I_0\cos^2\frac{\phi}{2}$$

A path difference of one whole wavelength corresponds to a phase difference of $2\pi$, so in general $\phi = \dfrac{2\pi}{\lambda}\times(\text{path difference})$.

Where the path difference is $\lambda$, $\phi = 2\pi$ and $\cos^2\pi = 1$, so the intensity is the maximum possible: $K = 4I_0$.

Where the path difference is $\lambda/3$, $\phi = 2\pi/3$, and

$$I = 4I_0\cos^2\frac{\pi}{3} = 4I_0\times\left(\frac12\right)^2 = I_0 = \frac{K}{4}$$

Question 10.6

A beam of light consisting of two wavelengths, $650\ \text{nm}$ and $520\ \text{nm}$, is used to obtain interference fringes in a Young’s double-slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength $650\ \text{nm}$. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? [Take the slit separation $d = 2\ \text{mm}$ and the distance to the screen $D = 1.2\ \text{m}$; see the note below.]

Note on the question. As printed in the book, this question gives neither the slit separation nor the distance to the screen, and without them no distance on the screen can be found. The answers printed at the back of the book need $D/d = 600$ — which is what $d = 2\ \text{mm}$ and $D = 1.2\ \text{m}$, the values this problem is usually set with, give. Both parts are worked with those values below; any pair with the same ratio gives the same answers.

Question 10.6 (a)

Solution. The $n$th bright fringe is at $x_n = n\lambda D/d$, and the third is $n = 3$. With $\lambda = 650\times10^{-9}\ \text{m}$ and $D/d = 1.2/(2\times10^{-3}) = 600$:

$$x_3 = 3\times(650\times10^{-9})\times600 = 1.17\times10^{-3}\ \text{m}$$

Question 10.6 (b)

Solution. Each wavelength makes its own set of bright fringes, and because the fringe width $\beta = \lambda D/d$ is proportional to $\lambda$, the two sets are spaced differently: $0.39\ \text{mm}$ apart for $650\ \text{nm}$ and $0.312\ \text{mm}$ apart for $520\ \text{nm}$. They coincide wherever the $n$th bright fringe of one lands on the $m$th bright fringe of the other:

$$n\,\frac{(650\ \text{nm})D}{d} = m\,\frac{(520\ \text{nm})D}{d} \quad\Longrightarrow\quad 650\,n = 520\,m \quad\Longrightarrow\quad \frac{n}{m} = \frac{4}{5}$$

The smallest whole numbers that work are $n = 4$ and $m = 5$: the fourth bright fringe of the longer wavelength falls on the fifth of the shorter, both at a path difference of $2600\ \text{nm}$. The distance is

$$x = 4\times(650\times10^{-9})\times600 = 1.56\times10^{-3}\ \text{m}$$

Common mistakes

• Question 10.1(b): changing the frequency when light enters the water. The frequency is set by the source and survives refraction unchanged; only the speed and the wavelength drop. Students who remember that “something” changes by the factor $n$ often divide the frequency instead of the wavelength.
• Question 10.1(b): multiplying the wavelength by $n$. Light slows down in water, and at a fixed frequency a slower wave has a shorter wavelength, so $\lambda = 589/1.33$, not $589\times1.33$. A wavelength of $783\ \text{nm}$ in water should look wrong at once.
• Question 10.2(b): calling the emerging wavefront spherical. A convex lens usually brings light to a focus, so students picture converging spherical wavefronts. With the source exactly at the focus the rays leave parallel, and a wavefront perpendicular to parallel rays is a plane.
• Question 10.3(b): saying violet travels faster. Violet photons carry more energy, which tempts students to think violet light moves faster. In glass the opposite is true: violet has the larger refractive index, which is why the prism bends it most, and a larger $n$ means a lower speed.
• Question 10.4: counting the fourth bright fringe as $n = 3$ or $n = 5$. The central bright fringe is $n = 0$, so the fourth bright fringe is four fringe widths from it. Mixing up bright and dark fringe counts gives $3.5$ or $5$ instead of $4$.
• Question 10.5: scaling the intensity with the path difference. Intensity is not proportional to path difference, so the answer is not $K/3$. It goes as $\cos^2(\phi/2)$, and both the square and the half-angle matter: a student who uses $\cos\phi$ instead gets $-K/2$, a negative intensity, which cannot be right.
• Question 10.6(b): looking for a common fringe width instead of a common position. The fringes coincide where $n\lambda_1 = m\lambda_2$, which needs the smallest whole numbers in the ratio $650 : 520 = 5 : 4$. Students who add the two fringe widths, or take their difference, get a distance where no two bright fringes meet.

Practise next

• Chapter 11, Dual Nature of Radiation and Matter — the photoelectric effect, which the wave picture of light built in this chapter cannot explain.
• Chapter 9, Ray Optics and Optical Instruments — refraction, refractive index and prisms from the ray point of view, the background to Questions 10.1 and 10.3.