A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye. (a) What is the magnification produced by the lens? How much is the area of each square in the virtual image? (b) What is the angular magnification (magnifying power) of the lens? (c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
Given: The size of each square is 1 mm 2 , the object distance is 9 cm and the focal length of magnifying glass is 10 cm .
a)
The lens formula is given as,
1f=1v−1u
Where, object distance is u, focal length of converging lens is fand the image distance is v.
By substituting the given values in the above expression, we get
1/10=1/v−1/−9
1/v=1/10−1/9= 9−10 /90
v=−90cm
Magnification is given as,
m=− v /u
By substituting the given values in the above expression,
m=−(−90/9)=10
Each side of the square is magnified 10 times. So, the area of each square in the image is given as,
A'=A×(10)2
Where, the area of each square is A .
By substituting the given values in the above expression, we get
A=1 mm 2 ×100 =100 mm 2 =1 cm 2
Thus, the area of each square in the image is 1 cm 2 .
b)
Magnifying power of the lens is given as,
m= d | u |
Where, d is the least distance for distinct vision.
By substituting the given values in the above expression, we get
m= 25 | −9 | = 25 9 =2.8
Thus, the magnifying power of the lens is 2.8.
(c) The magnification in (a) is not the same as the magnifying power in (b).
The magnitude of magnification produce by the magnifying glass is 10 while its magnifying power is 2 .The two quantities will be equal when the image is formed at the near point of the eye at 25 cm.
Given: The size of each square is 1 mm 2 , the object distance is 9 cm and the focal length of magnifying glass is 10 cm .
a)
The lens formula is given as,
1f=1v−1u
Where, object distance is u, focal length of converging lens is fand the image distance is v.
By substituting the given values in the above expression, we get
1/10=1/v−1/−9
1/v=1/10−1/9= 9−10 /90
v=−90cm
Magnification is given as,
m=− v /u
By substituting the given values in the above expression,
m=−(−90/9)=10
Each side of the square is magnified 10 times. So, the area of each square in the image is given as,
A'=A×(10)2
Where, the area of each square is A .
By substituting the given values in the above expression, we get
A=1 mm 2 ×100 =100 mm 2 =1 cm 2
Thus, the area of each square in the image is 1 cm 2 .
b)
Magnifying power of the lens is given as,
m= d | u |
Where, d is the least distance for distinct vision.
By substituting the given values in the above expression, we get
m= 25 | −9 | = 25 9 =2.8
Thus, the magnifying power of the lens is 2.8.
(c) The magnification in (a) is not the same as the magnifying power in (b).
The magnitude of magnification produce by the magnifying glass is 10 while its magnifying power is 2 .The two quantities will be equal when the image is formed at the near point of the eye at 25 cm.
