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Give Some Examples of Refraction?
Suppose, a beam of rays from a point object after refraction reaches our eyes in another medium. Now if the refracted rays are produced backwards they meet at a point. It seems that the refracted rays are diverging from the second point. Clearly, the second point is the image of the first point. The image of any point object is formed in the same way. Thus a complete image of the object is formed.
If the object is situated in a denser medium and is viewed from a rarer medium, it appears closer to the surface of separation. For example, if we look at a fish inside water in a pond it appears nearer to the surface than the actual position.
If the object is situated in a denser medium and is viewed from a rarer medium, it appears closer to the surface of separation. For example, if we look at a fish inside water in a pond it appears nearer to the surface than the actual position.
If the object is situated in a rarer medium and is viewed from a denser medium, it appears to move away from the surface of separation, e.g., our earth is surrounded by a thick atmospheric layer composed of different gases. Star light comes to our eyes through this atmosphere. Hence we are the observer on earth is in a denser medium whereas the stars are in vacuum i.e., in rarer medium. So, the actual position of a star is far behind from its normal viewing position.
Object in denser medium and eye in rarer medium: Let the refractive indices of the two media a and b be µ2 and µ1 respectively and µ2 > µ1 [Fig.], An object P situated in a is viewed from b.
The surface of separation of a and b is a plane surface. A ray of light from P incident perpendicularly at A proceeds along AB without changing its direction. Another oblique ray PC incident at C is refracted along CD. The refracted rays AB and CD when produced backwards meet at Q. So when the two refracted rays reach the eyes of the observer, it will appear as if the two rays are coming from Q. So Q is the virtual image of P. In this case the image apparently rises towards the surface of separation.
If the angle of incidence and the angle of refraction are i and r,
µ2 sin i = µ1 sin r or, \(\frac{\mu_2}{\mu_1}\) = \(\frac{\sin r}{\sin i}\)
For near normal viewing, points A and C are close enough. So sinθ ≈ tanθ
Hence, \(\frac{\mu_2}{\mu_1}\) = \(\frac{\sin r}{\sin i}\) = \(\frac{\tan r}{\tan i}\) = \(\frac{A C / A Q}{A C / A P}\) = \(\frac{A P}{A Q}\) ….. (1)
If the observer is in air then, µ1 = 1.
Putting µ1 = 1 and µ2 = µ (say) in equation (1) and writing
AP = u, AQ = v,
we have µ = \(\frac{A P}{A Q}\) = \(\frac{u}{v}\) …. (2)
If d is the real depth of the object then,
µ \(=\frac{d}{\text { apparent depth of the object }}\)
or, apparent depth of the object = \(\frac{d}{\mu}\)
Hence apparent displacement
x = PQ = AP – AQ = d – \(\frac{d}{\mu}\) = d(1 – \(\frac{1}{\mu}\))
Therefore, apparent displacement of an object depends on real depth (d) of the object and the refractive index (µ) of the denser medium.
Refractive index of water with respect to air is \(\frac{4}{3}\).
If an object immersed in water is observed vertically from above the water, then its apparent displacement
x = d(1 – \(\frac{1}{4 / 3}\)) = \(\frac{d}{4}\)
General case: Apparent depth of an object when viewed from air through successive media of thicknesses d1, d2, d3, …….,dn having refractive indices µ1, µ2, µ3, ……., µn respectively is
Object in rarer medium and eye in denser medium: Let the refractive indices of the two media a and b be µ2 and µ1 respectively and µ2 > µ1 [Fig.]. An object P is situated in the medium b and it is seen from the medium a.
The surface of separation of a and b is a plane surface. A ray of light from P is incident perpendicularly at the point A on the surface of separation and proceeds straight along AB through the medium a without changing its direction. Another oblique ray PC is incident at C and proceeds along CD after refraction. The refracted rays AB and CD when produced backwards meet at Q. So when the two refracted rays reach the eyes of the observer, it will appear to him that the two rays are coming from Q. So Q is the virtual image of P. In this case the image appears to move far-ther away from the surface of separation.
If the angle of incidence and the angle of refraction of the inci-dent ray at C are i and r respectively, then according to Snell’s law, ,
If the points A and C are very close to each other i.e., if the ray PC is not so oblique, then CP ≈ AP and CQ ≈ AQ
∴ \(\frac{\mu_2}{\mu_1}\) = \(\frac{A Q}{A P}\) …. (3)
If the rarer medium in which the object is situated in air then µ1 = 1.
putting µ1 = 1 and µ2 = µ (say) in equation (3) we get,
µ = \(\frac{A Q}{A P}\)
∴ The refractive index of the denser medium with respect to air apparent height of the object from
\(=\frac{\text { the surface of separation }}{\text { real height of the object from }}\)
the surface of separation
If the real height of the object, AP = d, then
µ \(=\frac{\text { apparent height of the object }(A Q)}{d}\)
or, AQ = µd ……. (4)
So the apparent displacement of the object = PQ = AQ – AP = µd – d = (µ – 1)d
General case: Apparent height of an object in air when viewed from a medium of refractive index µn through successive media of thicknesses d1, d2, ……., dn having refractive indices µ1, µ2, ……, µn respectively is
µ1d1 + µ2d2 + …… + µndn = \(\sum_{i=1}^n \mu_i d_i\)
and its apparent displacement is
(µ1 – 1)d1 + (µ2 – 1)d2 + ….. + (µn – 1)dn
= \(\sum_{i=1}^n\left(\mu_i-1\right) d_i\)
Let an object in a medium of refractive index µ1 be viewed from a medium of refractive index µ2. Then we have,
If object is situated in a comparatively denser medium, then µ1 > µ2.
In that case, apparent depth < real depth.
If object is situated in a comparatively rarer medium, then µ1 < µ2
In that case, apparent depth > real depth.
Image Formed by Oblique Incident Rays
In the last section we talked of almost normal viewing. For more oblique incidence, the apparent displacement is higher.
A point object O/ in an optically denser medium (say, water) is viewed from a rarer medium (say, air), at different angles [Fig.]. For different positions of the observer, the locus of the different positions of the image is a curved line. This curved line is called caustic curve. The curve has two parts. These two parts meet at a point O’, known as cusp. The image of an object at O when viewed vertically downward, is formed at O’.
Fig. shows how the images A’, B’, C’, etc. of different points A, B, C, etc. on the base of a vessel or tank containing water will appear to an observer located at a given position. It is evident that the image for normal incidence is at the lowest position. The other images go on rising up as the oblique rays from the base produce the images. The image of the base of the vessel will be a curved surface indicated by A’B’C’. With the increase of the distance of the base of the vessel from the eye, the curved surface appears to rise higher. So if an observer stands in a shallow pond having equal depth everywhere, it appears to him that the pond near his feet is the deepest.
Image of an Object under a Parallel Slab
ABCD is a parallel glass slab [Fig.]. Its thickness is d and refractive index µ. P is a point object placed in air under the surface AB of the slab. A ray of light PX normally incident on AB goes undeviated along XY. Another ray PQ incident obliquely is refracted along QR. After that the ray is further refracted along RS. Since the two faces AB and DC of the glass slab are parallel, the rays PQ and RS will be parallel [vide section 2.4]. So an observer from above the slab will see the image of P at P’. PP’ is the apparent displacement of the point object towards the observer.
The normal NQN1 at Q intersects P’R at M. The quadrilateral PP’MQ is a parallelogram.
∴ PP’ = QM
∴ The apparent displacement of the point object
= PP’ = QM = d(1 – \(\frac{1}{\mu}\))[see section 2.5]
So the apparent displacement of an object does not depend on the position of the object under the lower face of the glass slab. It only depends on the thickness of the slab (d) and the refractive index of its material (µ).
Some Examples of Refraction
A coin immersed in water: A coin is placed at the bottom of a pot such that the coin is just not visible [Fig.]. If eyes are set at same position and the pot is now filled with water, the coin becomes visible. Because the rays from P are refracted from denser to rarer medium and are bent away from the normal, it reaches our eyes. As a result the refracted rays appear to diverge from P’, i.e., the virtual image of the coin is formed at P’, situated above the coin.
A rod partly immersed in water: Let a straight rod be partly immersed in water. When the rod is held obliquely in water, the portion of the rod in water will appear to be bent upward [Fig.].
The reason is the same as above. The light rays coming from the portion immersed in water are refracted from denser to rarer medium, and hence bent away from the normal. So the point A of the rod appears to be raised at B. This happens for every point of the immersed portion of the rod.
Object and medium having approximately equal refractive index: An object becomes invisible when it is sur-rounded by a medium having nearly equal refractive index. Since the refractive indices of both of them are almost same, a negligible refraction does occur from their surface of separation, and for refraction bending of light is negligible i.e., it travels undeviated. As a result the surface of separation is not visible.
The refractive indices of glycerine and glass are almost equal. So when a glass rod is immersed in glycerine the rod is not visible.
Multiple images in a thick glass mirror: If an object placed in front a thick mirror silvered at the back surface is viewed from a slanting direction, a series of images are formed.
In the Fig., when ray PA is incident on the front face at the point A, a very small portion of the light is reflected along AK producing a faint image at P1. The remaining larger portion of light is refracted into the mirror along AB and is reflected back along BC from the silvered surface. A large portion of this reflected ray within the glass comes out along CL producing the second image P2 which is the brightest of all the images.
The remaining part of the ray CD is reflected back from the surface Y to the silvered surface where it is again reflected. The process continues and gradually fainter images are formed. The different images lie on the line drawn perpendicular to the surface X from the object P.
The ray BC, which is the first reflected ray from the silvered surface, is the brightest and it travels along CL. Consequently, image P2 is the brightest. Hence the brightest second image is considered to be the image of the object. More oblique the incident rays are, more will be the amount of reflection from the front surface of the mirror and brightness of the image P1 will increase accordingly.
Apparent thickness of thick glass mirror: in case of a water-filled bowl, the depth of the bowl appears to be less. Similarly, a thick glass mirror seems to be less thick than it really is.
= refractive index of glass = \(\frac{3}{2}\)
Therefore, apparent thickness of a mirror
= \(\frac{2}{3}\) × real thickness of the mirror
Numerical Examples
Example 1.
There is a mark at the bottom of a beaker. A liquid of refractive index 1.4 is poured into it. If the depth of the liquid is 3.5 cm then determine how much will the mark appear to rise when it is viewed from above?
Solution:
If the object is in denser medium and the observer is in rarer medium, refractive index of the denser medium relative to the rarer medium
\(=\frac{\text { real depth of the object }}{\text { apparent depth of the object }}\)
or, 1.4 \(=\frac{3.5}{\text { apparent depth of the object }}\)
or, apparent depth of the object = \(\frac{3.5}{1.4}\) = 2.5 cm
∴ Apparent upward displacement of the mark = 3.5 – 2.5 = 1 cm
Example 2.
There is a black spot at the bottom of a rectangular glass slab of thickness d and refractive index µ. When the spot is viewed perpendicularly from above, the spot appears to be shifted through a distance \(\frac{(\mu-1) d}{\mu}\) towards the observer. Prove it.
Solution:
Real depth of the black spot from the upper surface of the glass slab = d
Let the apparent depth of the black spot from the upper surface of the glass slab be d1
Refractive index of glass, µ = \(\frac{d}{d_1}\) or, d1 = \(\frac{d}{\mu}\)
∴ Apparent displacement of the black spot towards the observer
= d – d1 = d – \(\frac{d}{u}\) = d(1 – \(\frac{1}{\mu}\)) = d\(\left(\frac{\mu-1}{\mu}\right)\) (Proved)
Example 3.
In a beaker partly filled with water, the depth of water seems to be 9 cm. On pouring more water in it, the real depth of water is increased by 4 cm. Now the apparent depth of water seems to be 12 cm. Determine the refractive index of water and initial depth of water in the beaker.
Solution:
Let the refractive index of water be µ and initial depth of water in the beaker be x.
∴ µ = \(\frac{x}{9}\) or, x = 9µ
When more water is poured in the beaker, the real depth of water becomes (x + 4) cm.
∴ In the second case, µ = \(\frac{x+4}{12}\)
or, 12µ = x + 4 or, 12µ = 9µ + 4 or, 3µ = 4
∴ µ = \(\frac{4}{3}\) = 1.33
∴ Initial depth of water in the beaker
x = 9µ = 9 × \(\frac{4}{3}\) = 12 cm
Example 4.
A small air bubble exists inside a transparent cube of side 15 cm each. The apparent distance of the bubble observed from one face is 6 cm and from the opposite face its apparent distance becomes 4 cm. Determine the real distance of the bubble from the first face and refractive index of the material of the cube.
Solution:
Let the real distance of the bubble from the first face be x cm.
∴ The real distance of the bubble from the opposite face = (15 – x) cm
Let the refractive index of the material of the cube be µ.
We know if the object lies in a denser medium and eye in the rarer medium,
µ \(=\frac{\text { real distance }}{\text { apparent distance }}\)
∴ In the first case, µ = \(\frac{x}{6}\)
In the second case, µ = \(\frac{15-x}{4}\)
or, x = 9 cm and µ = \(\frac{9}{6}\) = 1.5
Therefore, the real distance of the bubble from the first face is 9 cm and refractive index of the material of the cube is 1.5.
Example 5.
A vessel is filled with two mutually immiscible liquids of refractive indices µ1 and µ2. The depths of the two liquids are d1 and d2 respectively. There is a mark at the bottom of the vessel. Show that the apparent depth of the mark when viewed normally is given by \(\left(\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}\right)\).
Solution:
The image of P is formed at Q due to refraction at the surface of separation B of the 1st and 2nd liquid. Another final image due to refraction in air from the second liquid is formed at R.
Example 6.
A rectangular slab of refractive index µ is placed on another slab of refractive index 3. Both the slabs are of the same dimensions. There is a coin at the bottom of the lower slab. What should be the value of µ such that when viewed normally from above, the coin appears to be at the surface of separation of the two slabs?
Solution:
Let thickness of each slab be d. According to the question the apparent depth of the coin = d.
∴ d = \(\frac{d}{\mu}+\frac{d}{3}\) or, 1 = \(\frac{1}{\mu}+\frac{1}{3}\)
or \(\frac{1}{\mu}\) = \(\frac{2}{3}\) or, µ = \(\frac{3}{2}\) = 1.5
Example 7.
A tank contains ethyl alcohol of refractive index 1.35. The depth of alcohol is 308 cm. A plane mirror is placed horizontally at a depth of 154 cm in it. An object is placed 254 mm above the mirror. Calculate the apparent depth of the image formed by the mirror.
Solution:
Depth of the mirror =1.54 m; object distance from the mirror = 0.254 m
The image of the object is formed at a distance of 0.254 m behind the mirror.
So the real depth of the image = 1.54 + 0.254 = 1.794 m
∴ Apparent depth of the image
\(=\frac{\text { real depth }}{\mu}\) = \(\frac{1.794}{1.35}\) = 1.33 m
Example 8.
A 20mmthick layer of water (µ = \(\frac{4}{3}\)) floats on a 35 mm thick layer of another liquid (µ = \(\frac{7}{5}\)) in a tank. A small coin lies at the bottom of the tank. Determine the apparent depth of the coin when viewed normally from above the water. [HS ’02]
Solution:
Real depth of the coin = d1 + d2 = 20 + 35 = 55 mm
∴ Apparent depth of the coin
= \(\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}\) = \(\frac{20}{\frac{4}{3}}+\frac{35}{\frac{7}{5}}\) = 15 + 25 = 40 mm
Example 9.
If a point source is placed at a distance of 18 cm from the pole of a concave mirror, its image is formed at a distance of 9 cm from the mirror. A glass slab of thick-ness 6 cm is placed between the point source and the mirror such that the parallel faces of the glass slab remains perpendicular to the principal axis of the mir-ror. If the refractive index of glass is 1.5 what will be the displacement of the image?
Solution:
Let P be the position of the object and in the absence of the glass slab, Q be the position of image formed by the concave mirror [Fig.].
Here, u = 18 cm,
OQ = v = 9 cm .
∴ According to \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\) we get,
\(\frac{1}{-9}\) + \(\frac{1}{-18}\) = \(\frac{1}{f}\) or, f = -6 cm
If the glass slab of thickness 6 cm is placed between the point source and the concave mirror, apparent displacement of the point source will take place towards the mirror. The rays coming from P appear to come from P’ after refraction.
Apparent displacement of P,
PP’ = d(1 – \(\frac{1}{\mu}\)) = 6(1 – \(\frac{1}{1.5}\)) = 2 cm
So in the second case, object distance u = -(18 – 2) = -16 cm; f = -6 cm; v = ?
we know, \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\) or \(\frac{1}{v}\) + \(\frac{1}{-16}\) = \(\frac{1}{-6}\)
or, \(\frac{1}{v}\) = \(\frac{1}{-6}\) + \(\frac{1}{16}\) = \(\frac{-10}{96}\) or, v = –\(\frac{96}{10}\) = -9.6 cm
∴ |OQ’| = 9.6
∴ Displacement of the image,
QQ’ = OQ’ – OQ = 9.6 – 9 = 0.6 cm
Example 10.
A plane mirror is made of glass having thickness 1.5 cm. Its back surface is coated with mercury. A man is standing at a distance of 50 cm from the front face of the mirror. If he looks at the mirror normally, where can he find his image behind the front face of the mirror? Refractive index of glass = 1.5 .
Solution:
Real depth of the mercury coated surface from the upper surface of the mirror =1.5 cm
If the apparent depth of the mercury coated surface is x cm, then \(\frac{1.5}{x}\) = 1.5 or, x = 1 cm x
So the mercury coated surface appears to be at a distance of 1 cm from the front face of the mirror.
Therefore, the distance of the man from the apparent position of the mercury coated surface = 50 + 1 = 51 cm
So the distance of the image from the apparent position of the mercury coated surface = 51 cm
∴ The distance of the image of the man from the front surface of the mirror = 51 + 1 = 52 cm.
Example 11.
An observer can see the topmost point of a narrow rod of height h through a small hole [Fig.]. ‘pie rod is placed inside a beaker. The beaker’s height is 3h and radius is h.
When the beaker is filled up to 2h of its height with a liquid the observer can see the entire rod. What is the value of the refractive index of the liquid? [IIT ’02]
Solution:
Let us assume that PQ is a thin straight rod kept in a beaker. B is a small hole on the wall of the beaker.
When the beaker is filled with a liquid up to a height 2h, then Q can be seen through the hole B.
Here, QD ray propagates through the liquid and gets refracted along DB in air.
According to Fig., ABRP is a square, where D is the midpoint of the diagonal PB.
Since the object is placed in the denser medium, so according to Snell’s law,
Therefore, required refractive index of the liquid is \(\sqrt{\frac{5}{2}}\).
Example 12.
A ray of light is incident and water at an angle of incidence i. If the ray finally emerges parallel to the surface of water as shown in the Fig., then what will be the value of µg ? [IIT 03]
Solution:
When refraction occurs due to the propagation of light rays from glass to water, we may write from Snell’s law,
µgsin i = µw sin r …. (1)
Again, when refraction occurs due to the propagation of light rays from water to air, we may write from Snell’s law,
µwsinr = µa sin90° = µa = 1 ……. (2)
From equations (1) and (2), we get
µg sin i = 1 or, µg = \(\frac{1}{\sin i}\)
Therefore, required refractive index of glass is \(\frac{1}{\sin i}\).