An object is placed (a) 6cm (b) 20cm from a thin convex lens of focal length 10cm. Determine the nature, position and magnification of the image in each case.
1/p + 1/q = 1/f
1/6 + 1/q = 1/10
M = -q/p
Upright or inverted is by the sign of q, + or -
To determine the nature, position, and magnification of the image formed by a thin convex lens, we can use the lens formula and magnification formula:
1. Nature of the image:
The nature of the image can be determined based on the position of the object with respect to the lens. If the image is virtual, erect, and diminished, it is considered to be a virtual image. If the image is real, inverted, and magnified, it is considered to be a real image.
2. Position of the image:
The position of the image can be determined using the lens formula:
1/f = 1/v - 1/u
where:
- f represents the focal length of the lens,
- v represents the image distance from the lens (positive for a real image, negative for a virtual image),
- u represents the object distance from the lens (positive if on the same side as the incident light, negative if on the opposite side).
3. Magnification of the image:
The magnification of the image can be determined using the magnification formula:
m = -v/u
where:
- m represents the magnification of the image,
- v represents the image distance from the lens,
- u represents the object distance from the lens.
Let's calculate the nature, position, and magnification of the image for both cases:
(a) When the object is 6cm from the lens:
Given:
f = 10cm
u = -6cm (negative because the object is on the opposite side of the incident light)
Using the lens formula:
1/10 = 1/v - (1/-6)
1/10 = 1/v + 1/6
6/(10v) = 7/6
10v = 36
v = 3.6cm
Using the magnification formula:
m = -v/u
m = -3.6/(-6)
m = 0.6 (negative value indicates inversion)
Therefore, for an object placed 6cm from the lens, the nature of the image formed is real, inverted, and magnification is 0.6. The image is located 3.6cm from the lens.
(b) When the object is 20cm from the lens:
Given:
f = 10cm
u = -20cm (negative because the object is on the opposite side of the incident light)
Using the lens formula:
1/10 = 1/v - (1/-20)
1/10 = 1/v + 1/20
2/(10v) = 3/20
10v = 40
v = 4cm
Using the magnification formula:
m = -v/u
m = -4/(-20)
m = 0.2 (negative value indicates inversion)
Therefore, for an object placed 20cm from the lens, the nature of the image formed is real, inverted, and the magnification is 0.2. The image is located 4cm from the lens.
Remember, positive values for image distances represent real images, while negative values represent virtual images. Inversions are indicated by negative magnification values.