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What should be the distance between the object in Exercise 9.23 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier? [Note: Exercises 9.22 to 9.24 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]
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1. Given Data
- Focal Length (f): +10 cm
- Object Area (Ao): 1 mm2
- Image Area (Ai): 6.25 mm2
2. Calculating Magnification (m)
The area magnification is the square of the linear magnification.
m2 = Ai / Ao
m2 = 6.25 / 1
m = √6.25
m = 2.5
3. Calculating Object Distance (u)
From the magnification formula (m = v/u), we have:
v = m × u
v = 2.5u
Using the Lens Formula:
1/v - 1/u = 1/f
Substituting v = 2.5u:
- 1/(2.5u) - 1/u = 1/10
- (1 - 2.5) / 2.5u = 1/10
- -1.5 / 2.5u = 1/10
- 2.5u = -15
- u = -15 / 2.5
- u = -6 cm
Required Object Distance = 6 cm
4. Can you see the squares distinctly?
No, you cannot see the squares distinctly.
Reason:
Let's calculate the position of the image (v):
v = 2.5 × u
v = 2.5 × (-6)
v = -15 cm
The virtual image is formed at a distance of 15 cm from the eye. However, the Least Distance of Distinct Vision (D) for a normal eye is 25 cm.
Since the image distance (15 cm) is less than the near point (25 cm), the eye cannot focus on it clearly, and the image will appear blurred.
Final Answer
- Object Distance: 6 cm
- Visibility: Not distinct (because image is closer than 25 cm).
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