A beam of light converges at a point P. Now a lens is placed in thepath of the convergent beam 12cm from P. At what point does thebeam converge if the lens is (a) a convex lens of focal length 20cm,and (b) a concave lens of focal length 16cm?
(a)
Given: The focal length of the convex lens is 20 cm , and the object distance from lens is 12 cm .
From the given situation, the object is virtual and image formed is real.
The lens formula is given as,
1 f = 1 v − 1 u
where, the focal length of the lens is f , the distance of the object from the lens is u , and the distance of the image from the lens is v .
By substituting the given values in the above expression, we get,
1 20 = 1 v − 1 12 1 v = 1 20 + 1 12 1 v = 3 + 5 60 v = 7.5 cm
Thus, the image formed is real and 7.5 cm away from the lens on its right side.
(b)
Given: The focal length of the concave lens is − 16 cm , and the object distance from lens is 12 cm .
The lens formula is given as,
1 f = 1 v − 1 u
where, the focal length of the lens is f , the distance of the object from the lens is u , and the distance of the image from the lens is v .
By substituting the given values in the above expression, we get,
1 − 16 = 1 v − 1 12 1 v = − 1 16 + 1 12 1 v = − 3 + 4 48 v = 48 cm
Thus, the image formed is real and 48 cm away from the lens on its right side.
(a)
Given: The focal length of the convex lens is
From the given situation, the object is virtual and image formed is real.
The lens formula is given as,
where, the focal length of the lens is
By substituting the given values in the above expression, we get,
Thus, the image formed is real and
(b)
Given: The focal length of the concave lens is
The lens formula is given as,
where, the focal length of the lens is
By substituting the given values in the above expression, we get,
Thus, the image formed is real and
