Guys Pls Solve this. ....
A object is placed 30 cm from the concave lens of focal length 15 cm. Find the nature, magnification and postion of the image. ??
Answers
Answered by
1
in concave lens f and u are always negative , so v will always negative and image will be imaginary.
1/v=1/f+1/u
=(1/-15)-(1/-30)
=(-2+1)/30
=-1/30
=> v=-30
m=-v/u=-(-30/-15)=-2
1/v=1/f+1/u
=(1/-15)-(1/-30)
=(-2+1)/30
=-1/30
=> v=-30
m=-v/u=-(-30/-15)=-2
01RoyalRajput:
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Thnq bro
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What is nature of the image
??
i have described,read carefully the lines
It is virtual or real
??
in concave lens f and u are always negative so v will always negative and image will always imaginary.
u and v both are negative ,so we have to show the interaction by dotted lines
Answered by
0
Answer:
m=-v/u
Explanation:
image =-2
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