The change in the speed of light as it moves from one medium to another is the fundamental reason for refraction. The amount of bending is related to the refractive index of the media, which is defined based on the speed of light in the medium.
Refraction of light, as it enters from one transparent medium to another, is due to
Asked in UPSC CDS 2 2021 · Topic: Optics (Reflection and Refraction at plane and spherical surfaces)
More Physics questions
- A concave mirror of radius of curvature 50 cm is used to form an image of an object kept at a distance of 25 cm from the mirror on its principal axis. What will be the position of the image from the mirror ?
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- When a ray of light travels from a denser medium to a rarer medium then
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- A ray of light strikes a glass slab (from air). Which one of the following is NOT correct?
- 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 ______.
- Concave mirror is used in headlights of vehicles because ______.
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