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At what distance from a concave lens of focal length 20cm a 6 cm tall object be placed so as to obtain it's image at 15cm from the lens? Also calculate the size of the image formed.?
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Given Data
- Focal length of the concave lens (f) = -20 cm (negative for concave)
- Height of the object (h) = 6 cm
- Image distance (v) = -15 cm (negative for virtual images)
Finding Object Distance
To find the object distance (u), we will use the lens formula:
1/f = 1/v + 1/u
Substituting the values:
1/-20 = 1/-15 + 1/u
Rearranging gives:
1/u = 1/-20 - 1/-15
To find a common denominator (60):
1/u = -3/60 + 4/60 = 1/60
Thus, u = 60 cm.
Conclusion on Object Placement
- The object should be placed at a distance of 60 cm from the concave lens.
Calculating the Size of the Image
To calculate the size of the image (h'), we can use the magnification formula:
Magnification (m) = h'/h = -v/u
Substituting the values:
m = -(-15)/(-60) = 15/60 = 1/4
Now substituting back to find h':
h' = m * h = (1/4) * 6 cm = 1.5 cm
Final Image Size
- The size of the image formed will be 1.5 cm tall.
Summary
- Object Distance: 60 cm
- Image Size: 1.5 cm tall
This setup allows for a virtual image that is smaller than the object, typical of concave lenses.
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At what distance from a concave lens of focal length 20cm a 6 cm tall object be placed so as to obtain it's image at 15cm from the lens? Also calculate the size of the image formed.?
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