the image produced by a concave mirror is one quarter the size of the object. if the object is moved 5cm closer to the mirror the image will only half the size of the object. Find the focal length of the mirror
Answers
Answer:
Explanation:
Suppose u = initial distance of object
magnification, m = 1/4
=> Now, Magnification of a lens: m = v/u
1/4 = v/u
v = u/4
=> By putting the value of v in mirror formula, we get
1/v + 1/u = 1/f
(-4 / u) + (-1/u) = 1/f ...(1)
=> Now, if the object is moved 5cm closer to the mirror the image will only half the size of the object.
Thus, u' = -(u - 5)
m' = 1/2
Thus, m' = v'/u'
1/2 =v' / -(u - 5)
v' = -(u - 5) / 2
=> By putting the value of v' in mirror formula, we get
1/v' + 1/u' = 1/f
(-2 / u-5) + (-1/ u-5) = 1/f ...(2)
From eq (1) and (2) ,
(4 / u) + (1/u) = (2 / u-5) + (1/ u-5)
5/u = 3/ u-5
5(u-5) = 3u
5u - 25 = 3u
2u = 25
u = 12.5
=> By substuting the value of u in eq (1), we get
(-4 / 12.5) + (-1/12.5) = 1/f
f = -2.5
here -ve sign indicates concave mirror
Thus, the focal length of the mirror is -2.5
Explanation:
ANSWER
In first case, if object distance is x, image distance =x/4.
While it is 2
nd
case, object distance becomes (x−5 cm) and image distance (x−5 cm)/2. Using mirror formula we get,
In first case, =
f
1
=−
x
5
In second case,
f
1
=−
(x−5)
3
Solving we get, x=12.5 cm and f=−2.5 cm and hence ∣f∣=2.5 cm