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Physics Topics can be both theoretical and experimental, with scientists using a range of tools and techniques to understand the phenomena they investigate.
What do you Mean by Total Reflecting Prism? Explain in Brief its Principle of Action
Types of Reflecting Prisms
Total reflecting prism: ‘Total internal reflection of light can easily take place in a prism made of crown glass and whose principal cross section is a right angled isosceles triangle. So this type of prism is called total reflecting prism.
ABC is the principal section of a total reflecting prism [Fig.]. Its sides AB and BC’ A
are equal and ∠ABC = 90°.
If the ray PQ is incident perpendicularly on the face AB, it is incident on the face AC at an angle of 45° which is greater than the critical angle of glass and air, about 42°. So it is totally reflected and passes along RST perpendicular to the side BC. Thus the hypotenuse face, AC of the prism acts as a plane mirror. So it is called a total reflecting prism.
It is to be noted that in this case the deviation of the ray is 90°.
Advantages of a total reflecting prism:
i) In case of a plane mirror multiple images may be formed due to several reflections from the front and the back surfaces of the mirror and the images thus formed is also not so bright. But in case of a total reflecting prism, only one bright image is formed due to total internal reflection.
ii) When the mercury coating of the back surface of a plane mirror is damaged, the image becomes indistinct. But no such problem arises in a total reflecting prism.
iii) In a plane mirror wastage of light occurs due to scattering but in a total reflecting prism there is no such wastage.
Disadvantages of a total reflecting prism:
- A total reflecting prism is more costly than a plane mirror.
- If the glass of the prism is not completely homogeneous, the image becomes less distinct.
Erecting prism: The inverted image of an object can be made erect with the help of a total reflecting prism.
Erecting of the image by deviating a ray through 0° : ABC is an isosceles right-angled prism, where ∠A = 90° and ∠B = ∠C = 45° [Fig.]. Suppose that QP is the inverted image of an object. The ray coming from Q after refraction on the face AB is incident on the face BC at an angle greater than the critical angle of glass and air. So the ray is totally reflected from this face and emerging from the face AC forms its image at point Q1. Similarly, a ray starting from P comes to the point P1. So, the prism forms the image P1Q1 of QP. Obviously P1Q1 is inverted with respect to QP, thus erecting an inverted object.
In this case, no deviation of the ray has taken place.
In instruments like telescope, binocular etc., total reflecting prism is therefore used to erect an inverted image.
Erecting of the image by deviating a ray through 180°: To get an erect image, the above mentioned prism can be used in another way.
QP is the inverted image of an object. The hypotenuse face BC of the prism ABC is held in front of it [Fig.]. A ray of light PX from P is Incident normally on the face BC and enters the prism. After refraction, It is incident at Y on the face AB. The angle of incidence of the ray XY at the face AB is 45° which is greater than the critical angle (nearly 42°) of glass and air. So the ray moves along YZ after total reflection at Y and is incident on the face AC. For the same reason the ray moves along ZR after total reflection at Z and is incident normally on the face BC. It emerges along RP’ and comes to the point P’.
Similarly, a ray coming from Q comes to the point Q’. Obviously according to the figure, the image P’Q’ becomes inverted relative to PQ. So the erect image P’Q’ of an inverted object is thus formed.
Each incident ray bends twice at 90°, thus producing a total deviation of 180°. So, an inverted image can be made erect by deviating a ray through 180°.
Prism periscope: Total reflecting prisms are commonly used nowadays in good quality periscopes instead of inclined parallel mirrors. This type of periscope is called prism periscope and the image formed by a prism periscope is more bright than that formed by a simple periscope.
The periscope tube contains two right-angled isosceles prisms P1 and P2 [Fig.]. P1 is fixed at the top in such a way that rays of light coming from a distant object enter the prism through a window and after total internal reflection goes downwards. The hypotenuse face P2, which is fixed at the bottom receives these rays and reflects them totally in a horizontal direction through the observation window. The observer thus sees an exact image of the distant object.
Numerical Examples
Example 1.
A man with a telescope can just observe the point A on the circumference of the base of an empty cylindrical vessel. When the vessel is filled completely with a liquid of refractive index 1.5, the man can just observe the middle point B of the base of the vessel without moving either the vessel or the telescope. If the diameter of the base of the vessel is 10 cm,what is the height of the vessel?
Solution:
When the vessel is empty, a light ray from the point A enters the telescope T following the straight path AO [Fig.]. When the vessel is filled with the liquid, a ray of light from the point B moves along BO and after refraction ¡n air enters the telescope. Let h be the height of the vessel.
Example 2.
A post of height 4 m is dipped straight in a pond. 1 m of the post remains above the water of the pond. If the rays of the sun are inclined at an angle of 45° to the surface of water what will be the length of the shadow of the post at the bottom of the pond? Refractive index of water µ = \(\frac{4}{3}\).
Solution:
ABC is the post and BFG is the surface of water [Fig.]. The length of the shadow at the bottom of the pond = CD.
From the figure,
∴ The length of the shadow of the post at the bottom of the pond = x + y = 1 + 1.876 = 2.876 m.
Example 3.
There is a point object at a height h above the surface of water In a tank. If the bottom of the tank acts as a plane mirror where will be the image formed? If an observer looks from air at the surface of water normally, calculate the distance of the image from the surface of water of the tank formed by the mirror-like bottom surface of the tank. Refractive index of water = \(\frac{4}{3}\).
Solution:
Q is a point source. For refraction in water the apparent position of Q is Q’ [Fig.].
PQ = h
Therefore the distance of Q’ from the bottom of the tank
= d + \(\frac{4}{3}\)h
So the image of Q’ will be formed at a distance of (d + \(\frac{4}{3}\)h) from the bottom of the tank.
The distance of the image from the surface of water
= d + d + \(\frac{4}{3}\)h = 2d + \(\frac{4}{3}\)h
If the apparent distance of the image from the surface of water to the eye of an observer in air medium is x then,
\(\frac{4}{3}\) = \(\frac{2 d+\frac{4}{3} h}{x}\)
or, x = \(\frac{3}{4}\left(2 d+\frac{4}{3} h\right)\) = \(\frac{3}{2}\)d + h
Example 4.
A rectangular glass slab of thickness 3 cm and of refractive Index 1.5 is placed in front of a concave mirror, perpendicular to its principal axis. The radius of curvature of the mirror is 10 cm. Where is an object to be placed on the principal axis so that its image will be formed on the object?
Solution:
The rectangular glass slab is placed perpendicular to the principal axis of the con- M,
cave mirror M1M2 [Fig.]. Suppose that if an object is placed at O on the principal axis, its image will be formed at O. OABM1 is the path of the ray. The ray after reflection at M1 retraces the path and forms image at O. So the ray BM1 must be incident on the mirror at M1 perpendicularly. If M1B is produced backward it intersects the principal axis at O’. So O’ is the centre of curvature of the concave mirror.
∴ PO’ = 10 cm
Now, OO’ = t(1 – \(\frac{1}{\mu}\)) = 3(1 – \(\frac{1}{1.5}\))
= 3 – 2 = 1 cm
So, the distance of O from the concave mirror
= PO’ + OO’ = 10 + 1 = 11 cm
Example 5.
A cross mark at the bottom of an empty vessel is focussed with the help of a vertical microscope. Now water (refractive index = \(\frac{4}{3}\)) is poured in the vessel. The height of water in the vessel is 4 cm. Another lighter liquid which does not mix with water and whose refractive index is \(\frac{3}{2}\), is poured on water. The height of the liquid is 2 cm. How much should the microscope he raised vertically to focus the mark again?
Solution:
The real depth of the combination of water and the liquid =4 + 2 = 6cm
The apparent depth of the cross mark
= \(\frac{4}{\frac{4}{3}}+\frac{2}{\frac{3}{2}}\) = 3 + \(\frac{4}{3}\) = 4.33 cm
So, to focus the cross mark, the microscope is to be raised vertically through a height (6 – 4.33) = 1.67 cm
Example 6.
A concave mirror of radius of curvature 1 m is placed at the bottom in a reservoir of water. When the sun is situated directly over the head, the mirror forms an
Image of the sun. If the depth of water is
(i) 80 cm and
(ii) 40 cm, calculate the image distances from the mirror. [Given, µw = \(\frac{4}{3}\)]
Solution:
The sun acts as an object situated at infinity. So its image will be formed by the concave mirror at its focus i.e., \(\frac{100}{2}\) = 50 cm above the mirror. When the depth of water in the reservoir is 80 cm the image is formed at a distance of 50 cm inside water from the mirror. But when the depth of water is 40 cm the image will be formed in air. Light rays will be refracted while passing from water to air. So, the refracted rays will converge at O’ and an image will be formed at O’ [Fig.].
Displacement of the image
= OO’ = t(1 – \(\frac{1}{\mu}\)) = 10(1 – \(\frac{3}{4}\))
= 10 × \(\frac{1}{4}\) = 2.5 cm
∴ distance of image from the mirror
= PO’= 50 – 2.5 = 47.5 cm
Example 7.
The width of a rectangular glass slab is 5 cm. From a point on its bottom surface light rays are incident on its top face and after total reflection form a circle of light of radius 8 cm. What is the refractive index of the glass slab’ [HS ‘15; WBCHSE Sample Question]
Solution:
Let O be a bright point at the bottom face of the rectangular slab. Light rays starting from O return to the bottom face after total reflection from the upper face of the slab. As a result, a circle of light of radius OA = OB = 8 cm is formed on the bottom face. So angle of incidence on the upper face is θc (critical angle) [Fig.].
According to the figure,
OP = \(\sqrt{O C^2+P C^2}\) = \(\sqrt{(4)^2+(5)^2}\) = \(\sqrt{41}\)cm
∴ refractive index,
µ = \(\frac{1}{\sin \theta_c}\) = \(\frac{1}{\frac{O C}{O P}}\) = \(\frac{O P}{O C}\) = \(\frac{\sqrt{41}}{4}\) = 1.6
Example 8.
Fig. shows a glass sphere having centre at O and two perpendicular diameters AOB and COD. A ray parallel to AOB is incident on the sphere at P where \(\overparen{A P}\) = \(\overparen{P C}\) and emerges from the sphere at B. Calculate the refractive Index of glass and the deviation of the emergent ray.
Solution:
According to the Fig., arc AP = arc PC
∴ ∠AOP = ∠POC = 45° = i,
∠POB = 45° + 90° = 135°
From the triangle POB,
r + r + 135° = 180° or, 2r = 45° or, r = 22.5°
∴ µ = \(\frac{\sin i}{\sin r}\) = \(\frac{\sin 45^{\circ}}{\sin 22.5^{\circ}}\)
= 2 cos 22.5° = 2 × 0.924 = 1.85
Angle of deviation,
δ = i – r + i – r = 2(i – r) = 2(45° – 22.5°)
= 2 × 22.5° = 45°
Example 9.
A glass slab is placed on a page of a book kept horizontally. What should be the value of the minimum refractive index of the glass slab so that the printed letters of the page will not be visible from any vertical side of the slab?
Solution:
It is assumed that there is a thin layer of air between the page of the book and the glass slab. So any ray coming from any portion of the page of the book is incident on the glass slab at an angle of almost 90° [Fig.].
If the angle of refraction is and refractive index of glass is µ, then, sin ϕ = \(\frac{1}{\mu}\)
If θc is the critical angle, then sin θc = \(\frac{1}{\mu}\). So here, ϕ = θc.
This refracted ray is incident on any vertical side of the slab at an angle θ (say). So θ + ϕ = 90°.
If θ is greater than ϕ, total internal reflection of light takes place and the printed letters of the page will not be visible from any vertical side. µ will be minimum when θ = ϕ
∴ 2ϕ = 90° or, ϕ = 45°
∴ μmin = \(\frac{1}{\sin \phi}\) = \(\frac{1}{\sin 45^{\circ}}\) = \(\sqrt{2}\) = 1.414
Example 10.
A surface of a prism having refractive index 1.5 is covered with a liquid of refractive index \(\frac{3 \sqrt{2}}{4}\). What should be the minimum angle of Incidence of an incident ray so that on the other surface of the prism the ray will be totally reflected from the surface covered with liquid? The refracting angle of the prism = 75° [sin 48°36’ = 0.75].
Solution:
Let the critical angle between the prism and the liquid be θc. If the ray of light is totally reflected from the surface covered with liquid then the angle of incidence of the ray on the surface is θc [Fig].
So, 1.5 sin θc = \(\frac{3 \sqrt{2}}{4}\) sin 90° = \(\frac{3 \sqrt{2}}{4}\)
or sin θc = \(\frac{3 \sqrt{2}}{4} \times \frac{10}{15}\) = \(\frac{1}{\sqrt{2}}\)
∴ θc = 45°
Again r1 + θc = A or, r1 + 45° = 75°, or r1 = 30°
Now, sin i1 = µsin r1 = 1.5 sin 30° = 0.75 or, i1 = 48°36′