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Read the following passage and answer the questions that follow :
Two students, A and B were performing an activity on refraction through glass slab and refraction through prism respectively. Student 'A' observed that the emergent light ray is parallel to the direction of incident ray, while student 'B' observed that the emergent ray is making a certain angle with respect to the direction of the incident ray in the prism, called angle of deviation. He also noted that the angle of deviation is different for different colours of light.
(a) When white light passes through a glass slab, it does not show dispersion, while it shows dispersion when it passes through a glass prism. Why ?
(b) Why do we get spectrum when white light passes through tiny water droplets present in air ?
(c) (i) For a given pair of mediums, if we change the angle of incidence gradually, will it change the speed of light in that medium ? Explain.
OR
(c) (ii) Calculate the speed of light in a given medium if angle of incidence in air is $60^\circ$ and angle of refraction in the medium is $30^\circ$. Given that speed of light in air is $3 \times 10^8\ \text{m/s}$.
Two students, A and B were performing an activity on refraction through glass slab and refraction through prism respectively. Student 'A' observed that the emergent light ray is parallel to the direction of incident ray, while student 'B' observed that the emergent ray is making a certain angle with respect to the direction of the incident ray in the prism, called angle of deviation. He also noted that the angle of deviation is different for different colours of light.
(a) When white light passes through a glass slab, it does not show dispersion, while it shows dispersion when it passes through a glass prism. Why ?
(b) Why do we get spectrum when white light passes through tiny water droplets present in air ?
(c) (i) For a given pair of mediums, if we change the angle of incidence gradually, will it change the speed of light in that medium ? Explain.
OR
(c) (ii) Calculate the speed of light in a given medium if angle of incidence in air is $60^\circ$ and angle of refraction in the medium is $30^\circ$. Given that speed of light in air is $3 \times 10^8\ \text{m/s}$.
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(a) Glass slab shows no dispersion because the colours dispersed by first
refracting surface recombine and emerge out as white light at the
second surface. (1)
In a glass prism different colours bend at different angles hence shows
dispersion
(b) Tiny water droplets act like prism. They refract and disperse the
sunlight to show spectrum. (1)
(c)
(i) With change in angle of incidence, angle of refraction will also change.
Hence refractive index remains constant. (1)
$\text{Refractive index} = \frac{\text{speed of light in air}}{\text{speed of light in given medium}}$
Hence speed of light in that medium will remain constant. (1)
OR
(c)
(ii) $\text{n} = \frac{\sin \text{i}}{\sin \text{r}}$
$\text{n} = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$ ($\frac{1}{2}$)
$\text{n} = \frac{\text{speed of light in air}}{\text{speed of light in medium}}$
$\sqrt{3} = \frac{3 \times 10^8}{\text{v}}$ ($\frac{1}{2}$)
$\text{v} = \frac{3 \times 10^8}{\sqrt{3}}$
$\text{v} = \sqrt{3} \times 10^8 \text{ m/s}$ ($\frac{1}{2}$)
refracting surface recombine and emerge out as white light at the
second surface. (1)
In a glass prism different colours bend at different angles hence shows
dispersion
(b) Tiny water droplets act like prism. They refract and disperse the
sunlight to show spectrum. (1)
(c)
(i) With change in angle of incidence, angle of refraction will also change.
Hence refractive index remains constant. (1)
$\text{Refractive index} = \frac{\text{speed of light in air}}{\text{speed of light in given medium}}$
Hence speed of light in that medium will remain constant. (1)
OR
(c)
(ii) $\text{n} = \frac{\sin \text{i}}{\sin \text{r}}$
$\text{n} = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$ ($\frac{1}{2}$)
$\text{n} = \frac{\text{speed of light in air}}{\text{speed of light in medium}}$
$\sqrt{3} = \frac{3 \times 10^8}{\text{v}}$ ($\frac{1}{2}$)
$\text{v} = \frac{3 \times 10^8}{\sqrt{3}}$
$\text{v} = \sqrt{3} \times 10^8 \text{ m/s}$ ($\frac{1}{2}$)