The intensity of light passing through a lens can be found using the formula I=Io - Io / csc^2 theta, where Io is the intensity of the light coming in, I is the intensity of the light that emerges or goes out of the lens, and theta is the angle between. a) Find and simply and equivalent expression. b) Suppose light passes through a polarized lens at angle 3.14/4 to the original lens. How much less intense than the original light will the light be now?

Question

The intensity of light passing through a lens can be found using the formula I=Io - Io / csc^2 theta, where Io is the intensity of the light coming in, I is the intensity of the light that emerges or goes out of the lens, and theta is the angle between. a) Find and simply and equivalent expression. b) Suppose light passes through a polarized lens at angle 3.14/4 to the original lens. How much less intense than the original light will the light be now?

Answer 1

I=Io-(I=/csc^2theta)

=Io-Io(sin^2theta)
=Io(1-sin^2theta)
=Io(cos^2theta)

For part b, assuming that it is the same type of light entering the lens, the intensity I entering should be the same. Then you can create two equations and use proportion to solve

Answer 2

a) To simplify the expression, we can use the trigonometric identity csc^2(theta) = 1 + cot^2(theta).

Substituting this into the given formula, we get:

I = Io - Io / (1 + cot^2(theta))

Rearranging the terms, we can simplify this as:

I = Io * (1 - 1 / (1 + cot^2(theta)))

b) Given that theta = 3.14/4, we can substitute it into our simplified expression:

I = Io * (1 - 1 / (1 + cot^2(3.14/4)))

To evaluate cot^2(3.14/4), we note that cot(theta) = 1 / tan(theta). So, cot(3.14/4) = 1 / tan(3.14/4). Since tan(theta) = sin(theta) / cos(theta), we have tan(3.14/4) = sin(3.14/4) / cos(3.14/4).

Using the values sin(3.14/4) = cos(3.14/4) = sqrt(2)/2, we can simplify cot(3.14/4) as:

cot(3.14/4) = 1 / tan(3.14/4) = 1 / (sqrt(2)/2) = 2/sqrt(2) = sqrt(2)

Now, substituting the value back into the expression for I:

I = Io * (1 - 1 / (1 + sqrt(2)^2))

Simplifying further:

I = Io * (1 - 1 / (1 + 2))

I = Io * (1 - 1/3)

I = Io * (2/3)

Therefore, the light will be 2/3 less intense than the original light.

Answer 3

a) To simplify the expression, we can start by expanding the right-hand side of the equation:

I = Io - (Io / csc^2(theta))

Since csc(theta) is equal to 1/sin(theta), we can rewrite the equation as:

I = Io - (Io / (1/sin^2(theta)))

Next, we can invert the fraction within the parentheses:

I = Io - (Io * (sin^2(theta)))

Now, simplify further by factoring out Io:

I = Io * (1 - sin^2(theta))

Using the trigonometric identity sin^2(theta) + cos^2(theta) = 1, we can substitute cos^2(theta) for 1 - sin^2(theta):

I = Io * cos^2(theta)

Therefore, an equivalent expression for the intensity of light passing through a lens is I = Io * cos^2(theta).

b) Given that the angle between the light passing through the polarized lens and the original lens is theta = 3.14/4, we can substitute this value into the simplified equation:

I = Io * cos^2(3.14/4)

Now, calculate cos(3.14/4) using a calculator or trigonometric table. The value of cos(3.14/4) is approximately 0.707.

Substituting this value back into the equation:

I = Io * (0.707)^2

Simplifying further,

I = (0.5 * Io)

Therefore, the light passing through the polarized lens will be less intense than the original light. It will be half as intense, specifically 0.5 times the intensity of the original light.