if a concave mirror has focal length 10 cm find thw two position where the object can be placed to give in each case an image twice the size of object
Answers
Given :-
▪ A concave mirror of focal length, f = - 10 cm.
To Find :-
▪ The two positions at which the object should be placed to obtain an image twice the size of the object.
Solution :-
Given that the image is twice the size of the object, we know
⇒ Magnification, m = - v/u = h₀ / h
Where,
- v = image distance
- u = object distance
- h₀ = image height
- h = object height
In case first, let's suppose the image we get is real, which means the distance will be negative.
For REAL image, magnification = negative
⇒ -(v) / u = -2
[ Since, Image is twice the size of the object ]
⇒ -v = -2u
⇒ v = 2u
⇒ 1/v = 1/2u ...(1)
For VIRTUAL image, magnification = positive
⇒ -(v) / u = 2
[ Since, The image is twice the size of the Object ]
⇒ -v = 2u
⇒ v = - 2u
⇒ 1/v = -1/2u ...(2)
CASE 1 (REAL IMAGE)
Using the mirror formula, we have
⇒ 1/f = 1/v + 1/u
⇒ -1/10 = 1/2u + 1/u [ from (1) ]
⇒ -1/10 = (1 + 2)/2u
⇒ -1/10 = 3/2u
⇒ -1/5 = 3/u
⇒ u = -15 cm
Case 2 (VIRTUAL IMAGE)
Similarly, using the mirror formula, we have
⇒ 1/f = 1/u + 1/v
⇒ -1/10 = 1/u - 1/2u [ from (2) ]
⇒ -1/10 = (2 - 1)/2u
⇒ -1/10 = 1 / 2u
⇒ 2u = -10
⇒ u = -5 cm
Hence, The two possible positions at which the object can be placed to obtain image twice the size of the object are -15 cm and -5 cm. Here, The negative sign indicates that the object must be placed infront of the mirror.