Refraction through a Rectangular Glass Block and Magnifying Glass: Refraction of light is seen in everyday life in the twinkling stars, advanced sunrise and delayed sunset, non-twinkling of planets etc. Optical devices like lenses can converge or diverge light rays passing through them because of refraction. It seems as if a pencil is broken when dipped in a beaker filled with water due to refraction. These phenomena are common in our surroundings, and we can spot many such occurrences due to refraction.

Have you ever tried to play with a Paperweight made of transparent glass?  Of course, you would have. But do you remember how things appear when seen through it?

A light ray incident at an angle going from a rarer medium to a denser medium bends towards the Normal. We verify this by doing some experiments on a rectangular glass slab.

Also, when we study a light ray travelling through a slab, a new term, Lateral shift, creates interest.

Most of the phenomena occurring in our everyday lives can be explained by light’s rectilinear motion. In this article, let’s study refraction through a glass slab, the lateral shift phenomenon and the working of a magnifying glass.

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Refraction of Light

The phenomenon of bending or deviation of the light ray when it passes from one medium to the other medium is called the refraction of light. This bending of light rays at the media boundary occurs due to the difference in speed of light in various media.

The laws of refraction of light:
1. The incident ray (IR), the refracted ray (RR) and the normal ray (NR) to the interface of the two transparent media at the point of incidence all lie in the same plane.
2. Snell’s law: The ratio of the sine of the angle of incidence to the angle of refraction \((r)\) is a constant called refractive index. According to this law,
\(\frac{{\sin \,i}}{{\sin \,r}} = {\text{constant}} …(1)\)
Where \(r = \) the angle of refraction The constant in equation \((1)\) is the refractive index of medium \(2\) w.r.t medium \(1\) denoted by \({n_{21}}.\)

Refractive Index

The refractive index of a medium depends on the nature of a medium, and it decides the speed of light in it. Refractive index is a scalar quantity, unit less and dimensionless quantity.

Types of Refractive Index

Absolute refractive index	Relative refractive index
(i) When a light ray travels from air to any denser medium, then the refractive index of a medium with respect to air is called its absolute refractive index i.e. \({}_{{\text{air}}}{\mu _{{\text{medium}}}} = \frac{C}{V}\)	(i) When a light ray travels from medium \(1\) to medium \(2\) then the refractive index of medium \(2\) with respect to medium \(1\) is called its relative refractive index i.e. \({}_1{\mu _2} = \frac{{{\mu _2}}}{{{\mu _1}}} = \frac{{{V_1}}}{{{V_2}}}\) (where \({{V_1}}\) and \({{V_2}}\) are the speed of light in medium \(1\) and \(2\) respectively).
(ii) Examples for absolute refractive index \({}_a{\mu _{{\text{glass}}}} = \frac{3}{2} = 1.5\) \({}_a{\mu _{{\text{water}}}} = \frac{4}{3} = 1.33\) \({}_a{\mu _{{\text{diamond}}}} = 2.4\) \({}_a{\mu _{{\text{crown}}}} = 1.52\) \({\mu _{{\text{vacuum}}}} = 1\) \({\mu _{{\text{air}}}} = 1.0003 \approx 1\)	(ii) Examples for relative refractive index (a) Suppose light enters from water to glass medium: \({}_w{\mu _g} = \frac{{{\mu _g}}}{{{\mu _w}}} = \frac{{\frac{3}{2}}}{{\frac{4}{3}}} = \frac{9}{8}\) (b) Suppose light enters from glass to diamond medium: \({}_g{\mu _D} = \frac{{{\mu _D}}}{{{\mu _g}}} = \frac{{2.4}}{{1.5}} = \frac{8}{5}\)

Note:
1. Cauchy’s equation: \(\mu = A + \frac{B}{{{\lambda ^2}}} + \frac{C}{{{\lambda ^4}}} + ….\) \(({\lambda _{{\text{Red}}}} > {\lambda _{{\text{violet}}}}\) so \({\mu _{{\text{Red}}}} < {\mu _{{\text{violet}}}})\)
2. If the light goes from medium \((1)\) to medium \((2),\) then \({}_1{\mu _2} = \frac{{{\mu _2}}}{{{\mu _1}}} = \frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{{{v_1}}}{{{v_2}}}\)

Refraction through a Glass Slab

A glass slab is an optical transparent medium made of glass with three dimensions: length, breadth, and height; Its shape is cuboidal.

Experiment

Take an opaque vessel and place a coin at the bottom. Move your head slowly away from the vessel till the coin disappears. Keep your head in the same position. Now, ask your friend to pour some water into the vessel. Observe the coin becoming visible. This phenomenon happens due to the bending of light.

These experiments show that the ray of light changes its direction when it travels from one medium to another. The phenomenon of bending of light rays when they pass from one medium to another is called the refraction of light. The phenomenon of refraction occurs in transparent media.

In the coin experiment, light travels straight when there is no water in the vessel. The coin was not visible initially. On adding water, the ray of light changed its direction at the water surface and reached the eye, and the coin appeared slightly higher than its original position. Thus, the coin appeared to be at a position higher than the initial one.

Lateral Shift Through a Glass Slab

On passing through a rectangular glass slab, a ray of light gets refracted twice at the two parallel surfaces, and the emergent ray becomes parallel to the incident ray. This is a unique property of a rectangular glass slab.

The refracting surfaces of a rectangular glass slab are parallel to each other. When a ray of light passes through a glass slab, it undergoes refracted twice at the two parallel faces of the glass slab. It emerges out parallel to its incident direction finally, i.e. the ray’s deviation is \(\delta = 0.\) The angle of emergence \(\left( e \right)\) is the same as the angle of incidence \(\left( i \right).\) The perpendicular (normal) distance between the incident and the emergent ray is the lateral shift of the ray, and it is given by- \(MN = t\,\sec r\sin \left( {i – r} \right)\)

Vertical Shift Through a Glass Slab

Observe this rectangular glass block carefully; it is placed in the path of a monochromatic light beam. You can see that because of refraction of light, the point of convergence or divergence appears to be shifted as shown-

Normal shift \(OO’ = x = \left( {1 – \frac{1}{\mu }} \right)t\)

Optical Path

The optical path is the path distance travelled by light in a vacuum at the same time in which it travels a given path distance in a medium.

	Time taken by a ray of light to pass through the medium \( = \frac{{\mu x}}{c};\) where \(x=\) geometrical path and \({\mu x=}\) optical path
	For two medium in contact optical path \( = {\mu _1}{x_1} + {\mu _2}{x_2}\)

Image Formation by a Lens

Lens	Location of the object	Location of the image	Nature of image
Magnification	Real virtual	Erect inverted
Convex	At infinity i.e. \(u = \infty \)	At focus i.e.\(v = f\)	\(m < 1\)diminished	Real	Inverted
Away from \(2f\) i.e. \(\left( {u > 2f} \right)\)	Between \(f\) and \(2f\) i.e.\(f < v < 2f\)	\(m < 1\)diminished	Real	Inverted
At \(2f\) or \(\left( {u = 2f} \right)\)	At \(2f\) i.e.\(\left( {v = 2f} \right)\)	\(m = 1\)same size	Real	Inverted
Between \(f\) and \(2f\) i.e. \(f < u < 2f\)	Away from \(2f\) i.e.\(v > 2f\)	\(m > 1\)magnified	Real	Inverted
At focus i.e. \(u = f\)	At infinity i.e. \(v = \infty \)	\(m = \infty \)magnified	Real	Inverted
Between optical centre and focus, \(u < f\)	At a distance greater than that of object \(v > u\)	\(m > 1\)magnified	Virtual	Erect

Note:
The minimum distance between an object and its real image formed by a convex lens is \(4f.\)
The maximum image distance for a concave lens is its focal length.

Magnifying Lens

While reading a book, we may have come across some pretty small words. Have you felt it difficult to read such small words? Let us look at a simple but exciting magical instrument that converts small words into big ones.

When we touch a glass tumbler lying on a table with our fingers, we see some spot on the glass after removing our fingers. Now, look at the same glass with a magnifying lens. What do we observe this time? We can also see our fingerprints on the glass. Hence, with the help of a magnifying lens, we can see those things which cannot be seen with our naked eyes.

The history of a simple magnifying lens goes like this. A man named Anton Van Leeuwenhoek used to work in a cloth store in Holland. To check the quality of the clothes, he counted the number of threads used in the cloth. Magnifying lenses can magnify objects up to 250 times of its original size. By combining the lenses, he made an even more powerful resolving instrument called the compound microscope. In modern times, the quality and the shape of the compound microscope are improving. These microscopes can enlarge an object many times more as compared to simple magnifying lenses. Due to its magnifying property, the compound microscope has been of great use for the scientific community, as it helped in discovering many microscopic organisms and bacteria that cause diseases. A compound microscope is an essential part of the medical field, especially for microbiologists, as it is used in various medical tests to identify the microorganisms in a patient’s body. Just as the lenses are useful in a microscope, they are also used in spectacles for improving vision, cameras, projectors and telescopes for seeing distant objects.

A simple microscope is a converging lens of a small focal length. When the object is held close to it, the lens produces a virtual, enlarged and erect image of the object.

The object is kept such that its image can be reversed comfortably, at \(25\,{\text{cm}}\) or more.
When the image is formed at the near point (when the object is held within focus)-

The linear magnification of the image \(m = \frac{v}{u} = v\left( {\frac{1}{v} – \frac{1}{f}} \right) = 1 – \frac{v}{f}\)
As \(v = \, – D\) we have \(m = 1 + \frac{D}{f}\)
Magnifying power \(\left( m \right)\) is also the ratio of the angle subtended by the image at the eye to that subtended by the object at the eye when placed at the least distance of distinct vision \(\left( D \right).\)
When the image is formed at infinity, i.e., when the object is kept at focus-
The angle subtended by the image at the eye \({\theta _i} = \frac{{h’}}{v} = \frac{h}{f}\) (\({\theta _i}\) is small) The angle subtended by the object at the eye when placed at a distance \(D\) is \({\theta _o} = \frac{h}{D}\)
Therefore the angular magnification produced
\(m = \frac{{{\theta _i}}}{{{\theta _o}}} = \frac{{\frac{h}{f}}}{{\frac{h}{D}}} = \frac{D}{f}\)
\(\therefore m = \frac{D}{f}\)
A simple microscope has a limited magnification \(\left( { \leqslant 9} \right)\) For larger magnifications, a compound microscope with two lenses, one enlarging the image of the other, is used.

Characteristics of a Simple Microscope

(i) It is a single bi-convex lens of lesser focal length.
(ii) It is also called a magnifying glass or reading lens.
(iii) The Magnification, when the final image is formed at \(D\) and \(\infty \) (i.e. \({m_D}\) and \({m_\infty }\))
\({m_D} = {\left( {1 + \frac{D}{f}} \right)_{\max }}\) and \({m_\infty } = {\left( {\frac{D}{f}} \right)_{\min }}\)

Note:
1. \({m_{\max }} – {m_{\min }} = 1\)
2. If the lens is kept at a distance \(a\) from the eye, then \({m_D} = 1 + \frac{{D – a}}{f}\) and \({m_\infty } = \frac{{D – a}}{f}\)

Solved Examples

Q1. The thickness ratio of two plates of the two transparent media, \(A\) and \(B,\) is \(6:4.\) If a ray of light passes through them in same time, then the refractive index of \(B\) with respect to \(A\) will be (a) \(1.5\)
(b) \(1.4\)
(c) \(1.75\)
(d) \(1.33\)
Solution: (a) On using the formula of the optical path, \(t = \frac{{\mu x}}{c}\)
\( \Rightarrow \frac{{{\mu _B}}}{{{\mu _A}}} = \frac{{{x_A}}}{{{x_B}}} = \frac{6}{4} \Rightarrow {}_A{\mu _B} = \frac{3}{2} = 1.5\)

Q2. A beam of light passes from air or vacuum into a medium of refractive index \(\mu .\) The angle of incidence is found to be two times that of the angle of refraction. Then the angle of incidence is given by
(a) \({\cos ^{ – 1}}\left( {\frac{\mu }{2}} \right)\)
(b) \(2{\cos ^{ – 1}}\left( {\frac{\mu }{2}} \right)\)
(c) \(2{\sin ^{ – 1}}\left( \mu \right)\)
(d) \(2{\sin ^{ – 1}}\left( {\frac{\mu }{2}} \right)\)
Solution: (b) By using Snell’s law for refraction: \(\mu = \frac{{\sin \,i}}{{\sin \,r}} \Rightarrow \mu = \frac{{\sin \,2r}}{{\sin \,r}} = \frac{{2\,\sin \,r\,\cos \,r}}{{\sin \,r}}\left( {\sin \,2\theta = 2\sin \,\theta \cos \,\theta } \right)\)
\( \Rightarrow r = {\cos ^{ – 1}}\left( {\frac{\mu }{2}} \right).\) So, \(i = 2r = 2{\cos ^{ – 1}}\left( {\frac{\mu }{2}} \right).\)

Q3. A rectangular slab having a refractive index \(\mu \) is placed over another slab of refractive index \(3,\) dimensions of both the slabs are identical. If a coin is placed below the lower slab, for what value of \(\mu \) will the coin appear to be placed exactly at the interface between the slabs when viewed from the top as shown?
(a) \(1.8\)
(b) \(2\)
(c) \(1.5\)
(d) \(2.5\)
Solution: (c)

The apparent depth of coin when seen from the top(rarer medium) \( = \frac{x}{{{\mu _1}}} + \frac{x}{{{\mu _2}}} = x\)
\( \Rightarrow \frac{1}{{{\mu _1}}} + \frac{1}{{{\mu _2}}} = 1 \Rightarrow \frac{1}{3} + \frac{1}{\mu } = 1 \Rightarrow \mu = 1.5\)

Q4. One face of a rectangular glass plate of thickness \(6\,{\text{cm}}\) is silvered. When an object is held \(8\,{\text{cm}}\) in front of the first face, it forms an image \(12\,{\text{cm}}\) behind the silvered face. What will be the refractive index of this glass?
(a) \(0.4\)
(b) \(0.8\)
(c) \(1.2\)
(d) \(1.6\)
Solution: (c)

As shown in the figure, the thickness of glass plate \(t = 6\,{\text{cm}}.\)
Let \(x\) be the apparent position of the silvered surface.
By the property of plane mirrors-
\(x + 8 = 12 + 6 – x \Rightarrow x = 5\,{\text{cm}}\)
Also \(\mu = \frac{t}{x} \Rightarrow \mu = \frac{6}{5} = 1.2.\)

Summary

1. The incident ray and the refracted ray are on the opposite sides of the normal at the point of incidence and are in different media.
2. A light ray incident at an angle going from a rarer medium to a denser medium bends towards the normal. In comparison, when it travels from a denser medium to a rarer medium, it bends away from the normal.
3. According to the law of refraction, “The incident ray, the refracted ray, and the normal at the point of incidence lie in one plane”.
4. Since for all media \(\mu > 1\) so, the optical path length \(\left( {\mu x} \right)\) is always greater than the geometrical path length \(\left( x \right).\)
5. In the refraction of light, frequency (and hence the colour) and phase do not change (while wavelength and velocity will vary).
6. In the refraction, the intensity of incident light decreases as it goes from one medium to another.
7. A transparent solid object is invisible in a liquid of the same refractive index (Because of no refraction).
8. Suppose a glass slab is kept over various coloured letters and seen from the top, the violet colour letters appear closer (Because \({\lambda _v} < {\lambda _R}\) so \({\mu _V} > {\mu _R}\) and from \(\mu = \frac{\lambda }{{\lambda ‘}}\) if \(\mu \) increases then ‘\(h\)’ decreases, i.e. Letter seems to be closer)

FAQ’s

Q1. What happens when light is refracted through a glass block?
A. Light waves change speed when they pass across the boundary between two substances with different densities, such as glass slabs. The speed of light increases when going into a rarer medium, and the ray bends away from the normal and causes them to change direction.

Q2. Why does light bend when it enters a glass block?
A. Light bends when it enters a glass block as rays of light slow down; when they enter the glass block, wave speed decreases, causing wavelength, in turn, to also decrease.

Q3. What happens to light if it enters a glass block along the normal?
A. If the light enters along the normal line, i.e. perpendicular (at \({90^ \circ }\)) to the block, then no refraction (change of direction) occurs. The light will be slowed down as it travels through the denser medium, but it will not change direction.

Q4. What is a good magnification for a magnifying glass?
A. A good magnification for a magnifying glass is 3 to 5.

Q5. Which light phenomenon happens in magnifying glass?
A. A magnifying glass makes objects appear much larger or magnified; this happens due to the phenomenon of refraction of light.

Q6. Is magnifying glass convex?
A. A magnifying glass is a bi-convex lens used to make an object appear much enlarged than it is. It works when the object is placed at a distance less than the focal length from the lens.

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