Question 2.9:
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:
Area of the house on the slide = 1.75 cm2
Area of the image of the house formed on the screen = 1.55 m2
= 1.55 × 104 cm2
Arial magnification, ma = Area of image / Area of object = ( 1.55 / 1.75 ) X 104
∴ Linear magnifications, ml = √ma
= √(1.55 / 1.75) X 104 = 94.11
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:
Area of the house on the slide = 1.75 cm2
Area of the image of the house formed on the screen = 1.55 m2
= 1.55 × 104 cm2
Arial magnification, ma = Area of image / Area of object = ( 1.55 / 1.75 ) X 104
∴ Linear magnifications, ml = √ma
= √(1.55 / 1.75) X 104 = 94.11
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