Notes on Lambert’s Cosine Law and Bunsen’s Photometer | Grade 11 > Physics > Photometry | KULLABS.COM

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#### Lambert’s Cosine Law

It states that when light falls obliquely on a surface, the illumination of the surface is directly proportional to the cosine of the angle θ between the direction of the incident light and the surface nurmal. The law is also known as the cosine emission law or Lambert's emission law. It is used to find the illumination of a surface when light falls on the surface along an oblique direction.

Suppose light from a source S falling on a surface area A as shown in the figure. The normal to the surface makes an angle$$\theta$$´ with the direction of light. Then, a component of A normal to the direction of the light ray is Acos$$\theta$$

The solid angle made by A at the source S,

\begin{align*} \Delta \omega &= \frac {\text {area}}{r^2} \\ &= \frac {A \cos \theta }{r^2} \\ \text {The total luminous flux passing normally through this surface area is } \\ Q &= L\Delta \omega \\ &= L\frac {A \cos \theta }{r^2} \\ \text {and illumination of the surface} \\ I &= \frac QA \\ &= \frac {L \cos \theta }{r^2} \dots (i) &= I_0 \cos \theta \\ \text {where} \: I_0 = L/r^2, \text {maximum illumination of surface. That is} \\ I &\propto \cos \theta \dots (ii) \end{align*}

which is Lambert’s cosine law, illumination of a surface is directly proportional to the cosine of the angle. Maximum illumination of a surface is obtained when light falls normally on the surface.
Hence illumination at a point due to a source is

1. directly proportional to the luminous intensity of the source
2. inversely proportional to the square of the distance of the point from the source and
3. directly proportional to the cosine to the cosine of the angle of incidence of luminous flux.

#### Bunsen’s Photometer

A Bunsen's grease spot photometer is used to compare the luminous intensities of the two source. If the luminous power of one source is known, than that of the second source can be calculated.

It consists of a piece of white paper with a grease spot at its centre and the two sources of light where luminous intensities are to be compared are placed on the two sides of the grease spot paper. These two sources are placed in such a way that the greased spot becomes indistinguishable from the paper when luminous flux per unit area coming from a white paper would be equal to that from the greased spot as shown in the figure. This grease spot transmits a fraction of incident light.

Suppose x represents the fraction of light transmitted by the grease-spot. Then the fraction of light reflected by it is (1-x). The light energy coming per second from a unit area of grease spot to the left the paper I1(1-x) + xI2 where I1 and I2 are the intensity of illumination of a left side and right side of the paper due to two sources S1 and S2 respectively. Similarly, light coming per second from the unit area of grease spot on the right side is I1x + (1-x)I2.

For the greased-spot to be distinguishable from the paper,

$$(1-x)I_1+xI_2=xl_1+(1-x)I_2$$

$$or ,I_1(1-2x)=(1-2x)I_2$$

$$or,I_1=I_2$$

$$we know I_1=\frac{L_1}{r_1^2}$$

$$andI_2=\frac{L_2}{r_2^2}$$

$$Then we have ,\frac{L_1}{r_2^2}=\frac{L_2}{r_2}$$

$$or\frac{L_1}{L_2}=\frac{r_1^2}{r_2^2}$$

Thus, by measuring r1 and r2 the luminous intensities of the two sources can be compared. If L1 is known, L2 can be calculated.

1,illumination at a point due to a source is

1. directly proportional to the luminous intensity of the source
2. inversely proportional to the square of the distance of the point from the source and
3. directly proportional to the cosine to the cosine of the angle of incidence of luminous flux.

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