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(B) Area of the house on the slide,$A_o = 1.75 \; cm^2$.Area of the image of the house on the screen,$A_i = 1.55 \; m^2$.Convert the area of the image

Vedclass · Accepted Answer

$94.11$

Vedclass · Answer

(B) Area of the house on the slide,$A_o = 1.75 \; cm^2$.Area of the image of the house on the screen,$A_i = 1.55 \; m^2$.Convert the area of the image to $cm^2$: $A_i = 1.55 	imes (100 \; cm)^2 = 1.55 	imes 10^4 \; cm^2$.The areal magnification $m_a$ is defined as the ratio of the area of the image to the area of the object: $m_a = \frac{A_i}{A_o} = \frac{1.55 	imes 10^4}{1.75}$.$m_a = 8857.14$.The linear magnification $m_l$ is the square root of the areal magnification: $m_l = \sqrt{m_a} = \sqrt{8857.14} \approx 94.11$.

The photograph of a house occupies an area of $1.75 \; cm^2$ on a $35 \; mm$ slide. The slide is projected onto a screen,and the area of the house on the screen is $1.55 \; m^2$. What is the linear magnification of the projector-screen arrangement?

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