# (solved)Question 2.9 of NCERT Class XI Physics Chapter 2

The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement.

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### Question 2.9 NCERT Class XI Physics

The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement.

### Solution in Brief

Areal magnification, i.e., zoom takes place when the slide is projected on a distant projector. Area is length-squared, so linear magnification will be square root of the areal magnification: 1.55 sq.m/1.75 sq. cm, that comes to 94.1 as the answer!

### Solution in Detail

Linear Magnification = $\displaystyle k$

area on slide is $\displaystyle l \times b$

area on screen is $\displaystyle (kl) \times (kb)$

dividing, $\displaystyle \frac{(kl)\times (kb)}{l \times b} = \frac{1.55 \text{ m}^2}{1.75 \text{ cm}^2}$

but $\displaystyle 1 \text{ m}^2 = 10^4 \text{ cm}^2\text{,}$

\begin{aligned} \implies k^2 &= \frac{1.55 \times 10^4 \text{ cm}^2}{1.75 \text{ cm}^2}\\\\ \implies k &= \sqrt{\frac{1.55 \times 10^4}{1.75}}\\\\ &= 100 \times \sqrt{\frac{155}{175}}\\\\ &= 100 \times \sqrt{\frac{31}{35}}\\\\ &\text{no calculators}\\ &\text{so do binomial approx}\\\\ &= 100 \times \bigg({1 - \frac{4}{35}}\bigg)^{\frac12}\\\\ &= 100 \times \bigg({1 - \frac12 \times \frac{4}{35}}\bigg)\\\\ &= 100 \times \frac{33}{35}\\\\ &= 94.28 \approx 94.3 \:\underline{Ans}\\ &\text{(3 significant figures}\\ & \text{in both 1.75 and 1.55}\\ & \text{so rounded to 3)} \end{aligned}

The assumption is that the magnification is isotropic - there is no distortion. So both the length and height of the image increase by the same factor of $\displaystyle k$.