(a) A lens of focal length 5 cm is being used by Debashree in the laboratory as amagnifying glass. Her least distance of distinct vision is 25 cm.(i) What is the magnification obtained by using the glass
Question
(a) A lens of focal length 5 cm is being used by Debashree in the laboratory as amagnifying glass. Her least distance of distinct vision is 25 cm.(i) What is the magnification obtained by using the glass
Solution
To find the magnification obtained by using the lens as a magnifying glass, we can use the formula:
Magnification (m) = - (v/u)
Where:
- v is the image distance
- u is the object distance
In this case, the lens is being used as a magnifying glass, so the object distance (u) is the least distance of distinct vision, which is given as 25 cm.
Now, we need to find the image distance (v). Since the lens is being used as a magnifying glass, the image formed is virtual and erect. In this case, the image distance is negative.
Using the lens formula:
1/f = 1/v - 1/u
Where:
- f is the focal length of the lens
Given that the focal length (f) of the lens is 5 cm, we can substitute the values into the lens formula:
1/5 = 1/v - 1/25
Simplifying the equation:
1/v = 1/5 + 1/25 1/v = (5 + 1)/25 1/v = 6/25
Cross-multiplying:
v = 25/6 cm
Since the image distance (v) is negative for a virtual image, we have:
v = -25/6 cm
Now, we can substitute the values of v and u into the magnification formula:
m = - (v/u) m = - (-25/6 cm) / (25 cm) m = 25/6
Therefore, the magnification obtained by using the lens as a magnifying glass is 25/6.
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