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Lunna [17]
3 years ago
14

An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2

50 cm from the object. What is the magnification?a)1.25b) 5.5c) 3.0d) 4.0e) 0.25
Physics
1 answer:
Leno4ka [110]3 years ago
7 0

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

M=-\frac{q}{p}

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

M=-\frac{200 cm}{50 cm}=-4.0

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q = - 93.334 nC

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GIVEN DATA:

Radius of ring  73 cm

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ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

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Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

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now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

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