A concave lens of focal length 20cm form an image 1/2 the size of the object. The object distance is
100cm
\( \frac{100}{q} \)
60cm
\( \frac{60}{7} \)
None of the above
Explanation
Video Explanation
No video available
Post your Contribution
Discussions (17)
Myschool I totally disagree with this please. The answer is suppose to be 20cm but since 20cm is not in the option, E is the answer. This is because for concave lens, the focal length is negative and also concave lens or diverging lens forms virtual images, so V should be negative too. If you solve it this way, your answer will be 20cm and E will be the correct option pls 
Greetings, The Myschool Team.
The correct answer is E(none of the above) and not C(60cm). The reason has to do with sign convention. Because the lens is a concave lens, its focal length is taken as negative. Also, the distances of virtual images are also taken as negative in all sign conventions. Since the image formed is virtual, its distance is also negative. Solving with these parameters gives an answer that is not in the options. Therefore, the correct answer is E(none of the above) and not C(60cm). Thank you.
20CM IS CORREt Hence option E is correct none of the option.
concave lens produces VED images i.e virtual erect and Diminished and virtual images takes negative. also for concave lenses, it is negative. hence
1/-20 = -2/u + 1/u
hence 20cm
my school abeg do official explanations to all this hard and stage acient years of past questions
I totally disagree with this answer the correct answer is supposed to be 20 cm since it is a concave lens where the focal length is negative and the lens is virtual
correct answer is -60cm since the focal length of a converging lens is taken as negative.
Since it a concave lens which is also a converging lens ..
The Focal Length will be positive = 20 cm
The image distance is positive regardless of the mirror
*While the image distance depends on the position, if it it opposite side from the object it will be positive but if it at same side as the object it will be negative*
. ————-(. )————
F. X
Since they said the image size is 1/2 of the size of the object
This means *V = 1/2U *
Using the mirror formula
1/F = 1/di + 1/dO
*1/20 = 1/0.5U + 1/U*
20 = 0.5U x U
——————
0.5U + U
20 = 0.5U^2
—————
1.5U
When a U cancels one U upward it becomes
20 = 0.5U/1.5
U = 20 x 1.5/ 0.5
U = 60cm ...
Hence the image will be 30cm from the Optical center of the lens
The magnification of the lens will be
M = V / U = 30/60
M = 0.5
Since the magnification is positive it will be real image , and since it less than 1 it will be deminished
*Hence it produces a real , inverted & Diminished image*
