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siniylev [52]
3 years ago
7

A converging lens of focal length 20 cm is used to form a real image 1.0 m away from the lens. How far from the lens is the obje

ct?
Physics
1 answer:
Galina-37 [17]3 years ago
6 0

Answer:

0.25 m

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem, we have

f = +20 cm=+0.20 m (the focal length is positive for a converging lens)

q = +1.0 m (the image distance is positive for a real image)

Solving the equation for p, we find

\frac{1}{p}=\frac{1}{f}-\frac{1}{q}=\frac{1}{0.20 m}-\frac{1}{1 m}=4 m^{-1}\\p=\frac{1}{4 m^{-1}}=0.25 m

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9) An object with a height of 18 cm is placed in front of a converging lens. The image has a
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Answer:

Explanation:

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b) Magnification = - image distance / object distance = -0.5

so image distance = 0.5 object distance

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1/6 = 1/(0.5 object distance) + 1/object distance

object distance = 18.0 cm

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A truck is carrying a refrigerator as shown in the figure. The height of the refrigerator is 158.0 cm, the width is 60.0 cm. The
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In order for the refrigerator not to tip over, the maximum acceleration of  1.86 m/s² must not be exceeded.

<h3>What is acceleration?</h3>

The term acceleration has to do with the rate at which velocity changes with time.

We have to take the moments at the tipping point of rotation as follows;

Clockwise moment = Anticlockwise moment

Hence;

F₂ * 1.58 m = F₁ * 0.67 m

The weight at half the width= 30 cm or 0.3 m

Height of refrigerator = 158 cm 0r 1.58 cm

Then;

m * a * 1.58 = m * 9.81 * 0.30

a = 1.86 m/s²

In order for the refrigerator not to tip over, the maximum acceleration of  1.86 m/s² must not be exceeded.

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Consider a concave mirror that has a focal length f. In terms of f, determine the object distances that will produce a magnifica
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We have that the magnification of each focal length is given respectively as

A) has u=3\frac{f}{2}

B) has u=4\frac{f}{3}

C) has  u=5\frac{f}{4}

From the question we are told that:

Focal Length F

Generally, the equation for Magnification is mathematically given by

M=\frac{-v}{u}

Therefore

v=2u

For A

M=-2

Therefore

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{u}+\frac{1}{2u}

Therefore

u=3\frac{f}{2}

For B

M=-3

Therefore

v=3u

Where

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{u}+\frac{1}{3u}

Therefore

u=4\frac{f}{3}

For C

M=-4

Therefore

v=4u

Therefore

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{u}+\frac{1}{4u}

Therefore

u=5\frac{f}{4}

Conclusion

From the calculations above we can rightly say that the magnifications of the values above are

A has u=3\frac{f}{2}

B has u=4\frac{f}{3}

C has  u=5\frac{f}{4}

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