A concave lens of focal length 60 cm is used to form an image of an object of length 9 cm kept at a distance of 30 cm from it. Use lens formula to determine the nature , position and length of image formed
Get 30 points for it
garimagill12:
Answer in brief please.
Not just method
Answers
Answered by
4
The image will be 60cm behind the mirror
it is erect,virtual and magnified.
Attachments:
where is your answer
The formula is 1/f=1/v-1/u
I have used the same formula
I did n't uploaded i came here to check my answer bt any ways ...bt now i sure my answer is far better than uuu
Look at the image first
okay I must be wrong ,sorry
It's ok
now thankyou
Ok
hey can you solve and send me this question
Answered by
1
Dear student,
● Answer -
a) Position of image is 20 cm to in front of lens.
b) Nature of image formed is real & inverted.
c) Length of image is 13.5 cm.
● Explanation -
# Given -
f = 60 cm
u = -30 cm
h1 = 9 cm
# Solution -
Using lens formula,
1/f = 1/v + 1/u
1/60 = 1/v + 1/-30
1/v = 1/60 + 1/30
v = 20 cm
Magnification by lens is calculated by -
M = -v/u = h1/h2
-20/-30 = 9/h2
h2 = 13.5 cm
a) Nature of image formed is real & inverted.
b) Position of image is 20 cm to in front of lens.
c) Length of image is 13.5 cm.
Hope this helps you...
● Answer -
a) Position of image is 20 cm to in front of lens.
b) Nature of image formed is real & inverted.
c) Length of image is 13.5 cm.
● Explanation -
# Given -
f = 60 cm
u = -30 cm
h1 = 9 cm
# Solution -
Using lens formula,
1/f = 1/v + 1/u
1/60 = 1/v + 1/-30
1/v = 1/60 + 1/30
v = 20 cm
Magnification by lens is calculated by -
M = -v/u = h1/h2
-20/-30 = 9/h2
h2 = 13.5 cm
a) Nature of image formed is real & inverted.
b) Position of image is 20 cm to in front of lens.
c) Length of image is 13.5 cm.
Hope this helps you...
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