Question 9 of 17intermediate🔧 ApplyNumerical2 marks

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.

Correct Answer

The linear magnification of the projector-screen arrangement is 94.1.

Given: Area of the house on the slide (AslideA_{slide}) = 1.75 cm21.75 \text{ cm}^2. Area of the house on the screen (AscreenA_{screen}) = 1.55 m21.55 \text{ m}^2.

Unit Conversion: To compare the areas, we convert the screen area into cm². 1.55 m2=1.55×104 cm2=15500 cm21.55 \text{ m}^2 = 1.55 \times 10^4 \text{ cm}^2 = 15500 \text{ cm}^2.

Calculation: Areal magnification is the ratio of the area of the image to the area of the object. Areal magnification=AscreenAslide=155001.75≈8857\text{Areal magnification} = \frac{A_{screen}}{A_{slide}} = \frac{15500}{1.75} \approx 8857.

Since area is proportional to the square of the linear dimension, the linear magnification (mm) is the square root of the areal magnification. m=Areal magnification=8857≈94.1m = \sqrt{\text{Areal magnification}} = \sqrt{8857} \approx 94.1.

Exercise: EXERCISES | Q: 1.9 | (Page 10)
For More Understanding

Explanation

To find the linear magnification, we first determine the areal magnification because the problem provides the areas. The areas must be in the same units, so 1.55 m21.55 \text{ m}^2 is converted to 1.55×104 cm21.55 \times 10^4 \text{ cm}^2. The areal magnification is the ratio of the image area to the object area. Since linear magnification relates to length and area has dimensions of length squared (L2L^2), the linear magnification is the square root of the areal magnification.

Solution Steps

  1. Step 1: Convert the area of the house on the screen from m2\text{m}^2 to cm2\text{cm}^2 to match the units of the slide area. 1.55 m2=1.55×104 cm2=15500 cm21.55 \text{ m}^2 = 1.55 \times 10^4 \text{ cm}^2 = 15500 \text{ cm}^2.

  2. Step 2: Calculate the areal magnification by dividing the area on the screen by the area on the slide. Areal magnification=155001.75≈8857\text{Areal magnification} = \frac{15500}{1.75} \approx 8857.

  3. Step 3: Determine the linear magnification by taking the square root of the areal magnification. m=8857m = \sqrt{8857}.

  4. Step 4: Solve for the final value. m≈94.1m \approx 94.1.

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