Optics Concept Page - 11

Result

No change in direction of light ray on refraction

From Snell's Law, we have
n1sinθ1=n2sinθ2$n_{1}sin{\theta }_1=n_{2}sin{\theta }_2$
When light ray is incident normally (i.e. i=0,i.e. θ1=0$i=0,i.e.{\theta }_1=0$), using Snell's law, we get, θ2=0${\theta }_2=0$. only speed changes and direction of  light remains the same.

Law

Principle of reversibility of light

According to principle of reversibility of light, if the path of the light is reversed after suffering a number of reflections and refractions, then it retraces its path. This means that if a light ray travels from medium 1 to medium 2 and has angle of incidence and angle of refraction as i$i$ and r$r$ respectively, then if the light is incident from medium 2 at an angle r$r$, then the angle of refraction in medium 1 will be i$i$.

Definition

Transmission of light from a denser medium to a rarer medium at different angles of incidence

When a ray of light enters from a denser medium to a rarer medium,it bends away from the normal, for example, the ray AO1B$A{O}_{1}B$ .The incident ray AO1$A{O}_1$ is partially reflected (O1C$O_{1}C$) and partially transmitted(O1B$O_{1}B$) or refracted, the angle of refraction (r ) being larger than the angle of incidence (i ). As the angle of incidence increases, so does the angle of refraction, till for the ray AO3$A{O}_3$, the angle of refraction is π2${\displaystyle \frac{\pi }{2}}$.The refracted ray is bent so much away from the normal that it grazes the surface at the interface between the two media. This is shown by the ray AO3D$A{O}_{3}D$ If the angle of incidence is increased still further (e.g.,the ray AO4$A{O}_4$ ), refraction is not possible, and the incident ray is totally reflected.

Definition

Apparent bending of a straight stick in water

This phenomenon occurs due to property of light called refraction of light.When immersing a stick in water, the rays of light pass from rarer medium to denser medium and they move towards the normal. So the part of stick immersed in water appears to be broken towards the normal drawn at the interface and gives an apparent bending. It also appears short like being raised up in the water(again the reason is the same).

Definition

Apparent depth and real depth

Real Depth is actual distance of an object beneath the surface, as would be measured by submerging a perfect ruler along with it.

Apparent depth in a medium is the depth of an object in a denser medium as seen from the rarer medium. Its value is smaller than the real depth.
Dapparent=Drealμ$D_{apparent}={\displaystyle \frac{D_{real}}{\mu }}$

Example

Calculate apparent depth of moving object

Problem:
A fish F, in the pond is at a depth of 0.8 m from the water surface and is moving vertically upward with velocity 2ms−1$2m{s}^{-1}$. At the same instant, a bird B is at a height of 6 m from the water surface and is moving downward with velocity 3ms−1$3m{s}^{-1}$. At this instant, both are on the same vertical line as shown in fig. Which of the following statements are correct?Solution:

height of B observed by F is 634+0.8=8.8${\displaystyle \frac{6}{{\displaystyle \frac{3}{4}}}}+0.8=8.8$

depth of F observed by B is 0.843+6=6.6${\displaystyle \frac{0.8}{{\displaystyle \frac{4}{3}}}}+6=6.6$

Example

Apparent depth of an object moved in the given medium

Example: A tank is filled with water to a height of 12.5 cm$12.5cm$. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm$9.4cm$. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63$1.63$ up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

Solution:
h1=12.5cm; h2=9.4cm$h_1=12.5cm;h_2=9.4cm$, μ=h1h2=1.33$\mu ={\displaystyle \frac{h_1}{h_2}}=1.33$
Now, water is replaced by μ1=1.63${\mu }_1=1.63$
So, μ1=hy${\mu }_1={\displaystyle \frac{h}{y}}$
⇒ y=7.67 cm$\Rightarrow y=7.67cm$
∴$\therefore$ distance by which the microscope should be moved up is 9.4−7.67=1.73 cm$9.4-7.67=1.73cm$

Example

Apparent depth of an Object

Example: A bird in air is at a height y$y$ from the surface of water. A fish is at a depth x$x$ below the surface of water. The refractive index of water is μ$\mu$. Find the apparent distance of fish from the bird.

Solution:
So, apparent distance of fish from the bird:
=yμair+xμwater$={\displaystyle \frac{y}{{\mu }_{air}}}+{\displaystyle \frac{x}{{\mu }_{water}}}$=y1+xμ$={\displaystyle \frac{y}{1}}+{\displaystyle \frac{x}{\mu }}$=y+xμ$=y+{\displaystyle \frac{x}{\mu }}$

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