What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in thefigure is to have an area of 6.25 mm2 . Would you be able to see thesquares distinctly with your eyes very close to the magnifier ?[Note: Exercises 9.29 to 9.31 will help you clearly understand thedifference between magnification in absolute size and the angularmagnification (or magnifying power) of an instrument.]
Given: The focal length of magnifying glass is 10 cm , size of each square is 1 mm 2 and the size of each square in virtual image is 6.25 mm 2 .
The linear magnification of an object is given as,
m = A A 0
Where, A the size of square in the virtual image and A 0 is the actual size of the square.
By substituting the given values in the above expression, we get
m = 6.25 1 = 2.5
Also,
m = v u
Where, v is the image distance and u is the object distance.
By substituting the given values in the above expression, we get
2.5 = v u v = 2.5 u (1)
The object distance ca be calculated using the lens formula.
1 f = 1 v − 1 u
Where, f is the focal length of the magnifying glass.
By substituting the given values in the above expression, we get
1 10 = 1 2.5 u − 1 u = 1 u ( 1 − 2.5 2.5 ) u = − 1.5 × 10 2.5 = − 6 cm
Substituting the value of u in equation (1), we get
v = 2.5 u = 2.5 × ( − 6 ) = − 15 cm
Since, the virtual image is formed at 15 cm , which is less than the near point of a normal eye. Thus, the virtual image cannot be seen distinctly.
Given: The focal length of magnifying glass is
The linear magnification of an object is given as,
Where,
By substituting the given values in the above expression, we get
Also,
Where,
By substituting the given values in the above expression, we get
The object distance ca be calculated using the lens formula.
Where,
By substituting the given values in the above expression, we get
Substituting the value of u in equation (1), we get
Since, the virtual image is formed at
