Question

According to ​Lambert's law​, the intensity of light from a single source on a flat surface at point P is given by Upper L equals k cosine squared (theta )where k is a constant. ​(a) Write L in terms of the sine function. ​(b) Why does the maximum value of L occur when thetaequals​0?

Answer 1

Answer:

(a) $L = k*(1 - sin^{2}(\theta))$        

(b) L reaches its maximum value when θ = 0 because cos²(0) = 1

Step-by-step explanation:

Lambert's Law is given by:

$L = k*cos^{2}(\theta)$   (1)

(a) We can rewrite the above equation in terms of sine function using the following trigonometric identity:

$cos^{2}(\theta) + sin^{2}(\theta) = 1$

$cos^{2}(\theta) = 1 - sin^{2}(\theta)$  (2)

By entering equation (2) into equation (1) we have the equation in terms of the sine function:

$L = k*(1 - sin^{2}(\theta))$        

(b) When θ = 0, we have:

$L = k*cos^{2}(\theta) = k*cos^{2}(0) = k$  

We know that cos(θ) is a trigonometric function, between 1 and -1 and reaches its maximun values at nπ, when n = 0,1,2,3...

Hence, L reaches its maximum value when θ = 0 because cos²(0) = 1.

I hope it helps you!

Answer 2

Answer:

$(a)L=k(1-sin^2\theta)\\(b)\text{0 is a critical point of L}$

Step-by-step explanation:

According to ​Lambert's law​, the intensity of light from a single source on a flat surface at point P is given by:

$L=kcos^2\theta, k is a constant$

(a)We are required to write L in terms of the sine function.

$cos^\theta+sin^2\theta=1\\cos^\theta=1-sin^2\theta\\L=k(1-sin^2\theta)$

(b)To obtain the maximum value of the function L, we examine its critical points.

$L=k-ksin^2\theta\\L'=2sin\theta cos\theta\\L'=sin 2\theta\\sin 2\theta=0\\2\theta=arcSin0\\2\theta=0\\\theta=0$

The maximum value of L occur when $\theta=0$ because 0 is a critical point of the function L.