REFRACTION ON PLANE SURFACES

REFRACTION ON PLANE SURFACES

Concepts of Refraction.

Refraction is the bending of a wave when it enters a medium where its speed is different. We can define it as the change in direction of a wave passing from one medium to another or from gradual change in the medium.

Causes of Refraction

i.                      The frequency of the refracted ray remains constant.

ii.                     Due to the partial reflection and absorption of light at the interface, the intensity of the refracted ray will be less than the incident ray.

iii.                    When the light crosses the boundary between two different media, deviation of light occurs resulting in refraction such that there is a change in wavelength and speed of light.

Examples are: I. A swimming pool seems much shallower than it actually is           II. a spoon appears bent when part of it is in water  

 III. a boy's legs look shorter when immersed in pool. All these effects are due to the refraction of light.

Laws of Refraction.

There are two laws governing the refraction of light:

I.                     The incident ray, the refracted ray and the normal at the point of incidence all lie on the same plane.

II.                    The ratio of the sine of the angle of incidence to the sine of angle of refraction is a constant for a given pair of media. The second law is known as Snell’s law. n = sin i / sin r (a constant) for a given pair of media.

If light is travelling from air to glass    ang = sine of angle of incidence on air / sin of angle of refraction on glass 

=speed of light in air / speed of light in glass

If light is travelling from glass to air    gna = sine of angle of incidence on glass / sine of angle of refraction in air

From principle of reversibility of light we have ang = 1/ gna



Refractive Index also called the index of refraction describes how fast light travels through the material. Refractive index is dimensionless. For a given material, the refractive index is the ratio between the speed of light in a vacuum and the speed of light in the medium.

Experimental proof of refraction (snell’s law)

 

Refractive Index From A Denser Medium To A Less Dense Medium

The depth of a river or a swimming pool always appears shallower than it actually is. When a glass block is placed on top of an object, e.g a pin or mark on a piece of paper, the object when view from directly above, appears nearer the top. This apparent depth is caused by refraction.

n = real depth / apparent depth

 Critical Angle And Total Internal Refraction.

The critical angle is the angle of incidence in the denser medium when the angle of refraction in the less dense medium is 90o.

Total internal reflection is the reflection of an incident ray of light at the interface between the medium of incidence and another medium of lower refractive index when the angle of incidence in the denser medium exceeds the critical angle.

Condition for total internal reflection

I Light must be travelling from an optically less dense medium.           II. The angle of incidence in the denser must be greater than the critical angle.

n = 1 / sin c where C is the critical angle

Example:
If the critical angle for a material is 42°. What is it’s refractive index?
n = 1 / sin c   = 1 / sin 42°   = 1.49
 Application Of Total Internal Reflection

( binocular, mirage, refractometer optical fibres)

 Refraction Through Prism.

When light rays pass through a rectangular glass block, the incident and the emergent rays are parallel to each other but the emergent ray is displaced to one side. There is no deviation or change in the direction of the emergent ray when compared with the incident ray.

Refractive Index Of Equilateral Triangle Prism

 n =sin 1/2 [A+   where A- the refracting angle of prism; D- minimum deviation; n- refractive index of a glass.



Example

A 60o glass prism has a refractive index of 1.63. Calculate the angle of (i) minimum deviation (ii) refraction of the light passing through the prism at minimum deviation and (iii) incident of minimum deviation.

Solution

A = 60oC, n= 1.63

i. n =  ;    1.63 =     ; 1.63 X sin 30 = sin ½ [60 + D] ;  sin-10.815= ½ [60 + D] ; 54.6 x 2 = 60 + D ; 109.2 -60 = D; D = 49.2

ii. 2r = A;  r= 60/2 = 30o

iii.  n =  ; sin I = n x sin r = 1.63 x sin 30; sin I = 1.63 x 0.5; I = sin-1 0.815; I = 54.6 

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