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An object of height 3cm is kept at a distance of 12 cm from a converging lens of power 10D. Find the position , nature and height of image?
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An object of height 3cm is kept at a distance of 12 cm from a convergi...
Given data:
Height of object (h_1) = 3 cm
Distance of object from lens (u) = -12 cm
Power of lens (P) = 10D
Finding the position of image:
Using the lens formula, 1/f = 1/v - 1/u
Given P = 1/f, we can find the focal length (f) of the lens by f = 1/P
Substitute P = 10D into the formula to get f = 1/10 = 0.1 m = 10 cm
Now, we can find the position of the image using the lens formula:
1/10 = 1/v + 1/12
1/v = 1/10 - 1/12
1/v = (6 - 5)/60
1/v = 1/60
v = 60 cm
Since the image distance is positive, the image is formed on the opposite side of the object.
Determining the nature of the image:
Since the image distance (v) is positive, the image is real and inverted.
Calculating the height of the image:
Using the magnification formula, M = -v/u
Substitute the values of v and u into the formula to get:
M = -60/-12 = 5
Now, we can find the height of the image (h_2) using the formula M = h_2/h_1
Substitute M = 5 and h_1 = 3 cm into the formula to get:
5 = h_2/3
h_2 = 5 * 3 = 15 cm
Therefore, the position of the image is 60 cm from the lens, the image is real and inverted, and the height of the image is 15 cm.
Height of object (h_1) = 3 cm
Distance of object from lens (u) = -12 cm
Power of lens (P) = 10D
Finding the position of image:
Using the lens formula, 1/f = 1/v - 1/u
Given P = 1/f, we can find the focal length (f) of the lens by f = 1/P
Substitute P = 10D into the formula to get f = 1/10 = 0.1 m = 10 cm
Now, we can find the position of the image using the lens formula:
1/10 = 1/v + 1/12
1/v = 1/10 - 1/12
1/v = (6 - 5)/60
1/v = 1/60
v = 60 cm
Since the image distance is positive, the image is formed on the opposite side of the object.
Determining the nature of the image:
Since the image distance (v) is positive, the image is real and inverted.
Calculating the height of the image:
Using the magnification formula, M = -v/u
Substitute the values of v and u into the formula to get:
M = -60/-12 = 5
Now, we can find the height of the image (h_2) using the formula M = h_2/h_1
Substitute M = 5 and h_1 = 3 cm into the formula to get:
5 = h_2/3
h_2 = 5 * 3 = 15 cm
Therefore, the position of the image is 60 cm from the lens, the image is real and inverted, and the height of the image is 15 cm.
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An object of height 3cm is kept at a distance of 12 cm from a converging lens of power 10D. Find the position , nature and height of image?
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An object of height 3cm is kept at a distance of 12 cm from a converging lens of power 10D. Find the position , nature and height of image? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about An object of height 3cm is kept at a distance of 12 cm from a converging lens of power 10D. Find the position , nature and height of image? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An object of height 3cm is kept at a distance of 12 cm from a converging lens of power 10D. Find the position , nature and height of image?.
An object of height 3cm is kept at a distance of 12 cm from a converging lens of power 10D. Find the position , nature and height of image? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about An object of height 3cm is kept at a distance of 12 cm from a converging lens of power 10D. Find the position , nature and height of image? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An object of height 3cm is kept at a distance of 12 cm from a converging lens of power 10D. Find the position , nature and height of image?.
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