Class 11 NCERT PHYSICS • Chapter 1: Units and Measurement • Question 9

QUESTION 9

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

SOLUTION

We need to calculate the linear magnification by comparing the house's area on the slide to its area on screen.

Concept: Areal magnification

The areal magnification ( $m_a$ ) is the ratio of the image area to the object area. For a two-dimensional projection, the linear magnification ( $m$ ) is the square root of the areal magnification.

m_a = \frac{A_i}{A_o} = m^2 \quad \implies \quad m = \sqrt{\frac{A_i}{A_o}}

Given:

Area of the house on the slide (object), $A_o = 1.75\,\text{cm}^2$
Area of the house on the screen (image), $A_i = 1.55\,\text{m}^2$

Solving:

Convert the object area from square centimeters to square meters to maintain consistent units:

\begin{aligned} A_o &= 1.75\,\text{cm}^2 \times \left(\frac{1\,\text{m}}{100\,\text{cm}}\right)^2 \\ &= 1.75 \times 10^{-4}\,\text{m}^2 \end{aligned}

Calculate the areal magnification ( $m_a$ ):

\begin{aligned} m_a &= \frac{A_i}{A_o} \\ &= \frac{1.55\,\text{m}^2}{1.75 \times 10^{-4}\,\text{m}^2} \\ &= \frac{15500}{1.75} \\ &= \frac{62000}{7} \end{aligned}

Determine the linear magnification ( $m$ ) by taking the square root of the areal magnification:

\begin{aligned} m &= \sqrt{\frac{62000}{7}} \\ &= \frac{20\sqrt{1085}}{7} \\ &\approx 94.11239 \end{aligned}

Answer

94.1