An object of height 2.5cm is placed at a distance of 15cm from the optical centre 'O' of a convex lens of focal length 10cm. Draw a ray diagram to find the position and size of the image formed. Mark optical centre 'O',principal focus F and height of the image on the diagram. My calculation for the position of image and height of the image is wrong but ray diagram is correct.How much mark should I expect to get from 2 in the board marking scheme?Plzzz answer .....
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In the ray diagram object size = AB = 2.5 cm, the focal length = CF =10 cm,
object distance =CB=u=15 cm, so the image at distance v is formed at: 1/f = 1/v − 1/u
or 1/10 = 1/v −1/−15
or 1/v = 1/10 −1/15 = 1/30
or v =30 cm
We know that I /O = v/u
so I / 2.5 = 30 / −15
or I = −2 × 2.5 = −5 cm
Therefore the image is formed by the convex lens at a distance of 30 cm on the other side of the lens which is real, inverted and magnified of the size 5 cm.
object distance =CB=u=15 cm, so the image at distance v is formed at: 1/f = 1/v − 1/u
or 1/10 = 1/v −1/−15
or 1/v = 1/10 −1/15 = 1/30
or v =30 cm
We know that I /O = v/u
so I / 2.5 = 30 / −15
or I = −2 × 2.5 = −5 cm
Therefore the image is formed by the convex lens at a distance of 30 cm on the other side of the lens which is real, inverted and magnified of the size 5 cm.
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In the ray diagram object size=AB=2.5cm, the focal length=CF=10cm, object distance =CB=u=15cm, so the image at distance v is formed at: 1f=1v−1uor 110=1v−1−15or 1v=110−115=130or v=30cmWe know that IO=vuso I2.5=30−15or I=−2×2.5=−5cmTherefore the image is formed by the convex lens at a distance of 30cm on the other side of the lens which is real, inverted and magnified of the size 5cm.
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