NCERT Solutions · Class 12 Physics Wave Optics

Speed of light in air \(\displaystyle c = 3\times10^{8}\ \mathrm{m/s}\), incident wavelength \(\displaystyle \lambda = 589\ \mathrm{nm}\), refractive index of water \(\displaystyle n = 1.33\).Frequency of the incident light (fixed by the source, unaffected by reflection or refraction): \[\nu = \frac{c}{\lambda} = \frac{3\times10^{8}}{589\times10^{-9}} = 5.09\times10^{14}\ \mathrm{Hz} \](a) Reflected light stays in the same medium (air), so its wavelength, frequency, and speed are all unchanged from the incident ray: \[\boxed{\lambda_{\text{refl}} = 589\ \mathrm{nm},\quad \nu_{\text{refl}} = 5.09\times10^{14}\ \mathrm{Hz},\quad v_{\text{refl}} = 3\times10^{8}\ \mathrm{m/s}} \](b) Refracted light: on entering water the frequency stays fixed (it is set by the source), but the speed changes to \(\displaystyle v = c/n\), and correspondingly the wavelength changes to \(\displaystyle \lambda' = \lambda/n\): \[v' = \frac{c}{n} = \frac{3\times10^{8}}{1.33} = 2.26\times10^{8}\ \mathrm{m/s} \] \[\lambda' = \frac{\lambda}{n} = \frac{589\ \mathrm{nm}}{1.33} = 442.9\ \mathrm{nm} \] Refracted light: \(\displaystyle \lambda' \approx 443\ \mathrm{nm}\), \(\displaystyle \nu' = 5.09\times10^{14}\ \mathrm{Hz}\) (unchanged), \(\displaystyle v' = 2.26\times10^{8}\ \mathrm{m/s}\).

A wavefront is the locus of points that a source's disturbance reaches simultaneously (a surface of constant phase), and by Huygens' construction its shape follows directly from the geometry of the source.(a) Light diverging from a point source: the disturbance spreads out equally in all directions at the same speed, so surfaces of constant phase are concentric spheres centred on the source — the wavefront is spherical.(b) Light emerging from a convex lens with a point source at its focus: a source at the focus produces rays that emerge parallel to the axis after refraction through the lens, and a bundle of parallel rays corresponds to wavefronts perpendicular to them everywhere — the emergent wavefront is plane.(c) The portion of a distant star's wavefront intercepted by Earth: the star's light forms spherical wavefronts of enormous radius (the star–Earth distance is astronomically large); the small patch of that sphere spanning the Earth's size is, to excellent approximation, flat — the intercepted wavefront is plane.

(a) Refractive index relates the speed of light in vacuum to the speed in the medium: \(\displaystyle n = c/v\), so \[v = \frac{c}{n} = \frac{3.0\times10^{8}\ \mathrm{m\,s^{-1}}}{1.5} = 2.0\times10^{8}\ \mathrm{m\,s^{-1}} \] Speed of light in glass \(\displaystyle = 2.0\times10^{8}\ \mathrm{m\,s^{-1}}\).(b) No — the refractive index of a medium is not the same for all wavelengths (this is dispersion); \(\displaystyle n\) increases as wavelength decreases, so violet light (shorter wavelength) has a higher refractive index than red light (longer wavelength) in glass. Since \(\displaystyle v = c/n\), a larger \(\displaystyle n\) means a smaller speed. Violet light travels slower than red light in a glass prism.

Young's double-slit fringe formula: the position of the \(\displaystyle n\)-th bright fringe from the central maximum is \[y_n = \frac{n\lambda D}{d} \] Given slit separation \(\displaystyle d = 0.28\ \mathrm{mm} = 0.28\times10^{-3}\ \mathrm{m}\), screen distance \(\displaystyle D = 1.4\ \mathrm{m}\), and the 4th bright fringe at \(\displaystyle y_4 = 1.2\ \mathrm{cm} = 1.2\times10^{-2}\ \mathrm{m}\) from the centre, solve for \(\displaystyle \lambda\): \[\lambda = \frac{y_4\, d}{nD} = \frac{(1.2\times10^{-2})(0.28\times10^{-3})}{4\times1.4} \] \[\lambda = \frac{3.36\times10^{-6}}{5.6} = 6.0\times10^{-7}\ \mathrm{m} \] \(\displaystyle \lambda = 600\ \mathrm{nm}\).

In Young's double-slit experiment the resultant intensity at a point where the path difference is \(\displaystyle \Delta\) (phase difference \(\displaystyle \phi = \frac{2\pi}{\lambda}\Delta\)) is \[I = I_{\max}\cos^{2}\!\left(\frac{\phi}{2}\right) \]At path difference \(\displaystyle \Delta = \lambda\): \(\displaystyle \phi = 2\pi\), so \(\displaystyle I = I_{\max}\cos^{2}(\pi) = I_{\max}\). This point is given to have intensity \(\displaystyle K\), so \(\displaystyle I_{\max} = K\).At path difference \(\displaystyle \Delta = \lambda/3\): \(\displaystyle \phi = \frac{2\pi}{\lambda}\cdot\frac{\lambda}{3} = \frac{2\pi}{3}\), so \[I = K\cos^{2}\!\left(\frac{\pi}{3}\right) = K\left(\frac{1}{2}\right)^{2} = \frac{K}{4} \] Intensity at path difference \(\displaystyle \lambda/3\) is \(\displaystyle K/4\).