Exercises — Wave Optics

1. 13 marksNCERT Exercises, Chapter 10

   Monochromatic light of wavelength 589 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.

   Hint. Frequency is fixed by the source and never changes at a boundary; decide which of the other two quantities must therefore change.

   The key principle is that frequency is set by the source and is unchanged by reflection or refraction. What changes on entering a new medium is the speed, and the wavelength adjusts to match.

   The frequency of the incident light is:

   $\nu =\frac{c}{\lambda }=\frac{3\times 10^8}{589\times 10^{-9}}=5.09\times 10^{14}\text{Hz}$

   (a) Reflected light. Reflection returns the light into the same medium, air, so nothing about it changes:

   $\lambda =589\text{nm},\nu =5.09\times 10^{14}\text{Hz},v=3\times 10^8{\text{m s}}^{-1}$

   (b) Refracted light. The frequency stays at $5.09\times 10^{14}$ Hz. The speed falls because water is optically denser:

   $v=\frac{c}{n}=\frac{3\times 10^8}{1.33}=2.26\times 10^8{\text{m s}}^{-1}$

   and the wavelength shortens in the same ratio:

   ${\lambda }^{\prime }=\frac{\lambda }{n}=\frac{589}{1.33}=443\text{nm}$

   The colour we perceive is tied to frequency rather than wavelength, which is why an object does not change colour when viewed underwater.

   ✦ (a) 589 nm, 5.09 x 10^14 Hz, 3 x 10^8 m/s (b) 443 nm, 5.09 x 10^14 Hz, 2.26 x 10^8 m/s

2. 23 marksNCERT Exercises, Chapter 10

   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.

   Hint. A wavefront is the surface joining points of equal phase; think about the symmetry of the source in each case.

   A wavefront is the locus of points that are all in the same phase, and its shape follows the symmetry of how the light is spreading.

   (a) Diverging from a point source. Light spreads out equally in every direction, so all points at the same distance are in phase. The wavefront is therefore spherical, expanding outwards as the light travels.

   (b) From a convex lens with the source at its focus. A source at the focal point produces a parallel emergent beam, because that is exactly what the focus means. Parallel rays correspond to a plane wavefront.

   (c) Light from a distant star. The star does emit spherical wavefronts, but by the time they reach the Earth the radius is astronomically large. A small portion of a sphere of enormous radius is indistinguishable from a flat surface, so the intercepted wavefront is effectively plane.

   Case (c) is the reason distant sources are treated as producing parallel light throughout optics.

   ✦ (a) Spherical and diverging (b) Plane (c) Plane, since a small part of a very large sphere is effectively flat

3. 33 marksNCERT Exercises, Chapter 10

   (a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is $3.0\times 10^8$ 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?

   Hint. Use the definition of refractive index for (a); for (b) recall that the refractive index of a medium depends on wavelength.

   (a) The refractive index is defined as the ratio of the speed of light in vacuum to that in the medium:

   $n=\frac{c}{v}⟹v=\frac{c}{n}=\frac{3.0\times 10^8}{1.5}=2.0\times 10^8{\text{m s}}^{-1}$

   (b) No, the speed is not independent of colour. The refractive index of a material varies with wavelength, being larger for shorter wavelengths. Since violet has a shorter wavelength than red:

   $n_{\text{violet}}>n_{\text{red}}⟹v_{\text{violet}}<v_{\text{red}}$

   So violet travels slower in the glass prism.

   This wavelength dependence is what splits white light into a spectrum, since violet is bent most and red least. Note that the phenomenon itself, dispersion, no longer has a section of its own in either this chapter or Chapter 9, though the effect is still needed to answer this question.

   ✦ (a) 2.0 x 10^8 m/s (b) No; violet travels slower than red, because the refractive index is larger for shorter wavelengths

4. 43 marksNCERT Exercises, Chapter 10

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

   Hint. The distance to the nth bright fringe is n times the fringe separation; rearrange for the wavelength.

   The position of the $n$th bright fringe measured from the central maximum is:

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

   Rearranging for the wavelength:

   $\lambda =\frac{x_{n}d}{nD}$

   Substituting $x_4=1.2\times 10^{-2}$ m, $d=0.28\times 10^{-3}$ m, $D=1.4$ m and $n=4$:

   $\lambda =\frac{1.2\times 10^{-2}\times 0.28\times 10^{-3}}{4\times 1.4}=\frac{3.36\times 10^{-6}}{5.6}=6.0\times 10^{-7}\text{m}$

   $\lambda =600\text{nm}$

   Note that the quantity $\lambda D/d$ is the separation between consecutive fringes, so this calculation is really measuring one fringe separation as $1.2/4=0.3$ cm. This chapter gives the fringe positions in Eq. 10.13 but never names or writes that separation as a formula of its own, so it is worth deriving it once and remembering it.

   ✦ lambda = 600 nm

5. 53 marksNCERT Exercises, Chapter 10

   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 $\lambda /3$?

   Hint. Convert each path difference into a phase difference, then use the intensity expression for two interfering waves of equal amplitude.

   For two coherent waves of equal amplitude the resultant intensity is:

   $I=4I_0{\cos }^2\left(\frac{\varphi }{2}\right)$

   where $\varphi$ is the phase difference. Path difference and phase difference are related by:

   $\varphi =\frac{2\pi }{\lambda }\times \text{path difference}$

   At path difference $\lambda$: $\varphi =2\pi$, so ${\cos }^2(\pi )=1$ and:

   $I=4I_0=K$

   This fixes the constant, giving $I_0=K/4$.

   At path difference $\lambda /3$: $\varphi =\frac{2\pi }{3}$, so:

   $I=4I_0{\cos }^2\left(\frac{\pi }{3}\right)=4I_0\times {\left(\frac{1}{2}\right)}^2=I_0$

   Substituting $I_0=K/4$:

   $I=\frac{K}{4}$

   The intensity falls to a quarter of the maximum, because a third of a wavelength of path difference corresponds to $120^{\circ }$ of phase, where the two amplitudes partially cancel.

   ✦ I = K/4

6. 64 marksNCERT Exercises, Chapter 10

   A beam of light consisting of two wavelengths, 650 nm and 520 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 nm. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

   Hint. Part (b) can be answered as a fringe order without any geometry; check carefully what data the question actually supplies for part (a).

   A note on the printed question. As it appears in this edition, the exercise supplies only the two wavelengths. It gives neither the slit separation $d$ nor the screen distance $D$, both of which are needed for any distance on the screen. Earlier editions stated $d=2$ mm and $D=1.2$ m. The results below are given first in symbolic form, which is complete for the question as printed, and then evaluated using that earlier data.

   (a) The third bright fringe for ${\lambda }_1=650$ nm lies at:

   $x_3=\frac{3{\lambda }_{1}D}{d}$

   With $d=2\times 10^{-3}$ m and $D=1.2$ m:

   $x_3=\frac{3\times 650\times 10^{-9}\times 1.2}{2\times 10^{-3}}=1.17\times 10^{-3}\text{m}=1.17\text{mm}$

   (b) Bright fringes coincide when the two systems have a maximum at the same place:

   $n_1{\lambda }_1=n_2{\lambda }_2⟹\frac{n_1}{n_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{520}{650}=\frac{4}{5}$

   The smallest whole numbers satisfying this are $n_1=4$ and $n_2=5$, so the fourth bright fringe of 650 nm falls on the fifth bright fringe of 520 nm. This part needs no geometry at all and is fully answerable as printed.

   The distance to that coincidence is:

   $x=\frac{4{\lambda }_{1}D}{d}=\frac{4\times 650\times 10^{-9}\times 1.2}{2\times 10^{-3}}=1.56\times 10^{-3}\text{m}=1.56\text{mm}$

   ✦ (a) 3 lambda D/d, which is 1.17 mm using the slit data omitted from this edition (b) the 4th bright fringe of 650 nm coincides with the 5th of 520 nm, at 4 lambda D/d, that is 1.56 mm

Solutions written by the tuition.in editorial team and checked against leph202.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.