A lens of focal length 12 cm produces a virtual image, whose linear dimensions are 3 that of the object. The distance of object and
the image from the lens are respectively
Answers
Given:
(i) A lens of focal length 12 cm produces a virtual image, whose linear dimensions are 3 that of the object
To find:
(i) The distance of object and the image from the lens are respectively
Solution:
We know that the lens formula is :
1/f = 1/u + 1/v
where,
f = Focal length of lens
v = Image distance
u = Object distance
On substituting value of focal length,f as 12cm, we get,
1/12 = 1/u + 1/v
⇒ 1/12 = (v + u)/uv
⇒ uv = 12(v + u)
⇒ 12v + 12u = uv ...(1)
From question, magnification is 3 times of the object.
Magnification = -v/u, since the image produced is virtual.
So, m = -v/u = 3
⇒ -v = 3u
⇒ v = -3u ...(2)
Replacing the value of 'v' from (2) in (1), we get,
12(-3u) + 12u = u(-3u)
⇒ -36u + 12u = -3u²
⇒ -24u = -3u²
⇒ 24u = 3u²
⇒ u² = 8u
⇒ u² - 8u = 0
⇒ u(u - 8) = 0
By Zero Product Rule, u = 0 and u = 8cm
∴ u = 8 cm
On substituting value of 'u' in equation (2), we get,
v = -3(8)
⇒ v = - 24 cm
The distance of object and the image from the lens are respectively are 8 cm and -24 cm respectively.
Given:
(i) A lens of focal length 12 cm produces a virtual image, whose linear dimensions are 3 that of the object
To find:
(i) The distance of object and the image from the lens are respectively
Solution:
We know that the lens formula is :
1/f = 1/u + 1/v
where,
f = Focal length of lens
v = Image distance
u = Object distance
On substituting value of focal length,f as 12cm, we get,
1/12 = 1/u + 1/v
⇒ 1/12 = (v + u)/uv
⇒ uv = 12(v + u)
⇒ 12v + 12u = uv ...(1)
From question, magnification is 3 times of the object.
Magnification = -v/u, since the image produced is virtual.
So, m = -v/u = 3
⇒ -v = 3u
⇒ v = -3u ...(2)
Replacing the value of 'v' from (2) in (1), we get,
12(-3u) + 12u = u(-3u)
⇒ -36u + 12u = -3u²
⇒ -24u = -3u²
⇒ 24u = 3u²
⇒ u² = 8u
⇒ u² - 8u = 0
⇒ u(u - 8) = 0
By Zero Product Rule, u = 0 and u = 8cm
∴ u = 8 cm
On substituting value of 'u' in equation (2), we get,
v = -3(8)
⇒ v = - 24 cm
The distance of object and the image from the lens are respectively are 8 cm and -24 cm respectively.