Object of height 6cm is placed at a distance of 20cm from a concave le...
Given that Object size, h = + 4 cm
Object distance, u = – 20 cm
Focal length, f = –10 cm
Image distance, v = ?Image size, h' = ?
1/f = 1/v + 1/u
or 1/v = 1/f - 1/u
= 1/(-10) - 1/(-20)
or 1/v = - (1/20)
v = -20 cm
The screen should be placed at 20 cm from the mirror. The image is real.
Also, magnification, m =h'/h = -(v/u) = -(-20)/(-20) = -1
The image is inverted and enlarged.
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Object of height 6cm is placed at a distance of 20cm from a concave le...
Given:
- Object height (h) = 6 cm
- Object distance (u) = -20 cm (negative sign indicates that the object is placed on the same side as the incident light)
- Focal length (f) = 10 cm
To find:
- Position of the image (v)
- Size of the image (h')
Formula:
The lens formula is given by:
1/f = 1/v - 1/u
Calculation:
Substituting the given values into the lens formula:
1/10 = 1/v - 1/-20
Simplifying the equation:
1/10 = 1/v + 1/20
Taking the LCM of the denominators:
1/10 = (1 + 2)/20
Simplifying further:
1/10 = 3/20
Cross multiplying:
20 = 10 * 3
Simplifying:
20 = 30
This is not possible, which means that there is no real image formed in this case. The image formed by the concave lens will be a virtual image.
Position of the image:
Since there is no real image formed, the position of the image cannot be determined.
Size of the image:
Since there is no image formed, the size of the image cannot be determined.
Conclusion:
When an object of height 6 cm is placed at a distance of 20 cm from a concave lens with a focal length of 10 cm, no real image is formed. Hence, the position and size of the image cannot be determined.