Question

❤❤Ahola❤❤
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♦ The photograph of a house occupies an area of 1.75 cm^2 on a 35 mm slide . The slide is projected on screen and the area of the house on the screen is 1.55m^2 .What is the linear magnification of project on screen ?
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Thnks ☺☺☺☺​

Answer 1

Given:

Area of the:

(i) House = $\sf{ {1.75 \: cm}^{2}}$

(ii) Image of the house (on screen) = $</strong><strong>\</strong><strong>sf</strong><strong>{</strong><strong> {1.55 \: m}^{2}</strong><strong>}</strong><strong> </strong><strong>$

$\boxed{\sf{1\:m = 100\:cm = {10}^{2} \: cm}}$

Now:

Area of the image of the house formed on the screen:

$\implies$ $\sf{1.55 \: {m}^{2}}$

$\implies$ $\sf{1.55 \times (10 ^{2} {cm})^{2}}$

$\implies$ $\sf{1.55 \times {10}^{4} {cm}^{2}}$

So:

$\boxed{\sf{Arial\:magnification = \frac{Area \: of \: image}{Area \: of \: slide}} }$

$\implies$ $\sf{ \frac{1.55 \times {10}^{4} {cm}^{2} }{ {1.75 \: cm}^{2} }}$

$\implies$ $\sf{0.8857 \times {10}^{4}}$

Therefore:

Linear mag = $\sf{\sqrt{(Arial \: magnification)}}$

$\implies$ $\sf{ \sqrt{(0.8857 \times {10}^{4} )}}$

$\implies$ $\sf{0.9411 \times {10}^{2}}$

$\implies$ $\sf{94.11}$

Final answer: 94.11

Answer 2

$\sqrt{(arial \: magnification)}$

$\sqrt{(arial \: magnification)}$$\sqrt{(0.8857 \times {10}^{4} )}$

$\sqrt{(arial \: magnification)}$$\sqrt{(0.8857 \times {10}^{4} )}$$0.9411 \times {10}^{2}$

$\sqrt{(arial \: magnification)}$$\sqrt{(0.8857 \times {10}^{4} )}$$0.9411 \times {10}^{2}$$94.11$