What is the Formula for Refraction at a Concave Surface?

Physics Topics can also be used to explain the behavior of complex systems, such as the stock market or the dynamics of traffic flow.

What is a Concave Refractive Surface of a Medium having Refractive Index?

i) When the object is real and lies in rarer medium and the image formed is virtual: Lët MPM’ be a concave surface separating two media of refractive indices μ₁ and μ₂ (μ₂ > μ₁) [Fig], Let P be the pole and C be the centre of curvature of the concave surface. A point object O is placed in the rarer medium on the principal axis OP. An incident ray OA, after refraction at the point A on the surface bends towards the normal AN and goes along AB in the denser medium. Another ray OP moving along the principal axis is incident on the surface normally and hence gets undeviated into the denser medium. The two refracted rays AB and PQ being divergent meet at I when produced
backwards. I is the virtual image of O.

Let angle of incidence, ∠OAC = i; angle of refraction, ∠BAN opposite angle ∠IAC = r; ∠ACO = θ; object distance, PO = -u; image distance, PI = -u; radius of curvature, PC = -R.

Now from the triangle AOC have,
\(\frac{\sin i}{C O}\) = \(\frac{\sin \theta}{A O}\) or, \(\frac{\sin i}{\sin \theta}\) = \(\frac{C O}{A O}\) ….. (1)
From the triangle AIC, we have
\(\frac{\sin r}{C I}\) = \(\frac{\sin \theta}{A I}\) or, \(\frac{\sin r}{\sin \theta}\) = \(\frac{C I}{A I}\) ….. (2)
Again considering refraction at the point A according to Snell’s law we have,
µ₁i = µ₂sin r

Equation (4) is called Gauss’ equation for refraction at a concave spherical surface.
If the object O is in air, µ₁ = 1 and µ₂ = µ (say), then
equation (4) becomes
\(\frac{\mu}{v}-\frac{1}{u}\) = \(\frac{\mu-1}{R}\) …. (5)

ii) When the object is virtual and the image formed is real: Let MPM’ be a concave surface separating two media of refracting indices µ₁ and µ₂(µ₂ > µ₁) [Fig]. Let P be the pole and C be the centre of curvature of the concave surface. O is the virtual point object on the principal axis of the concave surface and I is its real image.
Let angle of incidence ∠DA C = opposite angle ∠NAO = i; angle of refraction, ∠NAI = r; ∠ACP = θ; virtual object distance, PO = u; real image distance, PI = u; radius of curvature, PC = -R.

Now from ΔAOC we have,
\(\frac{\sin \left(180^{\circ}-i\right)}{C O}\) = \(\frac{\sin \theta}{A O}\) or, \(\frac{\sin i}{C O}\) = \(\frac{\sin \theta}{A O}\)
or, \(\frac{\sin i}{\sin \theta}\) = \(\frac{C O}{A O}\) …. (6)
From ΔACI we have,
\(\frac{\sin \left(180^{\circ}-r\right)}{C I}\) = \(\frac{\sin \theta}{A I}\) or, \(\frac{\sin r}{\sin \theta}\) = \(\frac{C I}{A I}\) …. (7)
Again considering refraction at the point A, according to Snell’s law we have,

When the object is real and lies in denser medium and the image formed is virtual: In Fig. MPM’ is a spherical surface which is concave towards the rarer medium.
The object O is placed in the denser medium. The virtual image of the object is I.
Here µ₂ > µ₁
Let angle of incidence, ∠DAO = i;
angle of refraction, ∠BAC = r; ∠ACP = θ
object distance, Po = -u;
image distance, PI = -v;
radius of curvature, PC = R


Again considering refraction at the point A, according to Snell’s law we have,
μ₂sini = μ₁sinr
or μ₂\(\frac{\sin i}{\sin \theta}\) = μ₁\(\frac{\sin r}{\sin \theta}\) …. (13)
∴ From the equation (11), (12) and (13) we have,

1. If object is real and lies in rarer medium, then relation \(\frac{\mu_2}{v}\) – \(\frac{\mu_1}{u}\) = \(\frac{\mu_2-\mu_1}{R}\) is valid irrespective of the type of the spherical refracting surface.

2. If object is real and lies in the denser medium, then rela-tion
\(\frac{\mu_1}{v}-\frac{\mu_2}{u}\) = \(\frac{\mu_1-\mu_2}{R}\) is valid irrespective of the type of the spherical refracting surface.

Here u = object distance, v = image distance, r = radius of curvature of spherical refracting surface, μ₁ = refractive index of rarer medium and μ₂ = refractive index of denser medium.

Numerical Examples

Example 1.
There is a small air bubble Inside a glass sphere (μ = 1.5) of radius 10 cm. The bubble is 4 cm below the surface and is viewed nearly normal from the out side. Find the apparent depth of the bubble.
Solution:
Here, u = -4 cm; r = -10 cm; μ₂ = 1.5; μ₁ = 1
O is the position of the bubble and I is the position of the image of the bubble [Fig.]

Thus the bubble will appear 3 cm below the top point of the sphere.

Example 2.
A point of red mark on the surface of a glass sphere is observed straight, nearly along the diameter from the opposite surface of the sphere. If the diameter of the sphere be 20 cm and refractive index of the glass be 1.5, find the position of the image.
Solution:
Let P be a point of red mark on the glass sphere being observed from the point A [Fig].
According to the question, object distance = AP = u = -20 cm; radius of curvature = r = -1o cm.

In this case, the object lies in the denser medium µ₂ = 1.5 and the observer is situated in the rarer medium (µ₁ = 1).
So, in case of refraction in the spherical surface BAC,

So, a virtual image will he formed on Q on the extended line AOP at a distance of 40 cm from the point A. So the virtual position of the red spot will be found (40 – 20) or 20 cm behind from its real position while looking through the sphere.

Example 3.
A mark exists at a distance of 3 cm on the axis from the plane surface of a hemisphere of glass. If the mark is observed from above the curved surface determine the apparent position of the mark. Radius of the hemisphere = 10 cm; refractive index of glass = 1.5.
Solution:
A is the position of the mark [Fig]. A’ is the position of its image.
u = -OA = -(OC – AC) = -(10 – 3) = -7 cm
r = -10 cm; µ₂ = 1.5 and µ₁ = 1

∴ Position of the image from the plane surface is at a distance of (10 – 6.09) or 3.91 cm.

Example 4.
A parallel beam of light travelling in water is refracted by a spherical air bubble of radius 2 mm situated in water. Find the position of the image due to refraction at the first surface and the position of the final image. Refractive Index of water = 1.33. Draw a ray diagram showing the positions of both the images.
Solution:
The ray diagram is shown in the Fig.
Let C be the centre of the spherical air bubble. P₁ and P₂ are the poles of the spherical surfaces. A beam of light parallel to the diameter of the sphere, after refraction at the first surface forms a virtual image I₁. After that it forms another virtual image I₂ due to refraction at the second surface.

For refraction at the first surface of the bubble (from water to air)
μ₁ = 1, μ₂ = 1.33; u = ∞ and r = 2mm
∴ \(\frac{1}{v}\) – \(\frac{1.33}{\infty}\) = \(\frac{1-1.33}{2}\) or, \(\frac{1}{v}\) = \(\frac{-1}{6}\) or, v = -6 mm
The negative sign indicates that the image I₁ is virtual and forms at 6 mm from the surface of the bubble on the water side. The refracted rays (which seem to come from I₁) are incident on the farther surface of the bubble. For this refraction, μ₁ = 1, μ₂ = 1.33, r = 2 mm and u = -(6 + 4) = -10 mm

The negative sign shows that the image is formed on the air side at 5 mm from the second refracting surface.

Measuring from the centre of the bubble the first image is formed at (6 + 2) or 8 mm from the centre and the second image is formed at (5 – 2) or 3 mm from the centre. Both images are formed on the side from which the incident rays are coming.

Example 5.
A spherical surface of radius of curvature R separates air (refractive Index = 1.0) from glass (refractive Index = 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O and PO = OQ. Find the distance of the object from the spherical surface.
Solution:
Let PO = OQ = x. Suppose object and image distances are u and u respectively.
We know, \(\frac{\mu_2}{v}-\frac{\mu_1}{u}\) = \(\frac{\mu_2-\mu_1}{R}\)
Here, µ₂ = 1.5, µ₁ = 1, v = +x, u = -x
From equation (1),
\(\frac{1.5}{x}-\frac{1}{-x}\) = \(\frac{1.5-1}{+R}\)
or, \(\frac{2.5}{x}\) = \(\frac{0.5}{R}\) or, x = 5R
Hence distance of the object from the spherical surface is 5R.