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Masja [62]
3 years ago
9

How far from a converging lens with a focal length of 16 cm should an object be placed to produce a real image which is the same

size as the object? Express your answer using two significant figures.
Physics
1 answer:
Feliz [49]3 years ago
4 0

Answer:

32 cm

Explanation:

f = focal length of the converging lens = 16 cm

Since the lens produce the image with same size as object, magnification is given as

m = magnification = - 1

p = distance of the object from the lens

q = distance of the image from the lens

magnification is given as

m = - q/p

- 1 = - q/p

q = p                                    eq-1

Using the lens equation, we get

1/p + 1/q = 1/f

using eq-1

1/p + 1/p = 1/16

p = 32 cm

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Prove(show) ''T=2π√(l/g)''​
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Answer:

Time period for Simple pendulum, T=2\pi\sqrt{\frac{l}{g}

Explanation:

The Simple Pendulum

Consider a small bob of mass m is tied to extensible string of length l that is fixed to rigid support. The bob is oscillating in the plane about verticle.

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Vertical component of the force on bob, F=-mg\sin\theta

Negative sign shows that its opposing the motion of bob.

Taking \theta as very small angle then, \sin\theta\sim\theta

F=-mg\theta    

Let x is the displacement made by bob from its mean position ,

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so, F=-mg\frac{x}{l}                ........(1)

Since, pendulum is in hormonic motion,

as we know, F=-kx

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F=-m\omega^2x                   .........(2)

From equation (1) and (2)

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T=2\pi\sqrt{\frac{l}{g}}

6 0
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