The speed of light (v), frequency (f), and wavelength (λ) are related by the equation v = fλ. When light enters a denser medium like glass from air, its speed (v) decreases, and since the frequency (f) remains constant, the wavelength (λ) must decrease proportionally (λ = v/f).
A ray of light strikes a glass slab (from air). Which one of the following is NOT correct?
Asked in UPSC Geoscientist 2021 · Topic: Optics (Reflection and Refraction at plane and spherical surfaces)
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Answer: D. The frequency of light in glass is less than that in air
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- An electromagnetic wave $\vec{E} = \vec{E}_0 e^{i k_0(3x+4y)-\omega t}$, travelling in air is incident on the x - z plane of a glass slab. The glass slab has a refractive index of 1.5. Which one among the following is the correct direction of propagation of the transmitted electromagnetic wave ? C. $0.4\hat{i} + \sqrt{21}\hat{j}$ D. $0.4\hat{i} - \sqrt{21}\hat{j}$ Ans: A - The incident electromagnetic wave is given by $ \vec{E} = \vec{E}_0 e^{i k_0(3x+4y)-\omega t}$. The wave vector $\vec{k}_i$ is the vector part multiplying $\vec{r}=(x,y,z)$ in the exponent (ignoring $k_0$). So, the incident wave vector in air is proportional to $3\hat{i} + 4\hat{j}$. We can write $\vec{k}_i = k_0 (3\hat{i} + 4\hat{j})$. - The magnitude of the incident wave vector is $|\vec{k} _i| = k_0 \sqrt{3^2 + 4^2} = 5 k_0$. In air, the refractive index $n_1 = 1$, so $|\vec{k}_i| = n_1 (\omega/c) = \omega/c$. This implies $k_0 = \omega/(5c)$. - The wave is incident on the x-z plane, which is the plane where $y=0$. The normal to this plane is along the $\hat{j}$ direction. Let's assume the glass slab is in the region $y>0$, and the air is in $y 0$, so it is travelling towards the interface $y=0$ from the region $y0$), the y-component of the transmitted wave vector must be positive (pointing into the glass region). So $k_{ty} = \frac{3\sqrt{21}}{2} k_0$. - The transmitted wave vector is $\vec{k}_t = 3k_0 \hat{i} + \frac{3\sqrt{21}}{2} k_0 \hat{j}$. The direction of propagation is given by the unit vector in the direction of $\vec{k}_t$. - The direction is proportional to $3\hat{i} + \frac{3\sqrt{21}}{2} \hat{j}$. - Let's check the options. Option A is $0.4\hat{i} + 0.2\sqrt{21}\hat{j} = \frac{2}{5}\hat{i} + \frac{\sqrt{21}}{5}\hat{j}$. - Check if $(3, \frac{3\sqrt{21}}{2})$ is proportional to $(\frac{2} {5}, \frac{\sqrt{21}}{5})$. The ratio of x-components is $3 / (2/5) = 15/2$. The ratio of y-components is $(\frac{3\sqrt{21}}{2}) / (\frac{\sqrt{21}}{5}) = \frac{3\sqrt{21}}{2} \times \frac{5}{\sqrt{21}} = \frac{15}{2}$. - Since the ratios are equal, the vectors are proportional. The direction of propagation of the transmitted wave is proportional to 182. The Black Board seems black because it ______.
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