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MArishka [77]
4 years ago
13

A diverging lens (f1 = -12.0cm) is located 50.0 cm to the left of a converging lens (f2 = 34.0 cm). A 2.0 cm-tall object stands

to the left of the diverging lens, exactly at its focal point. Determine the distance of the final image relative to the converging lens in centimeters.
Physics
1 answer:
pishuonlain [190]4 years ago
5 0

Answer:

The final image relative to the converging lens is 34 cm.

Explanation:

Given that,

Focal length of diverging lens = -12.0 cm

Focal length of converging lens = 34.0 cm

Height of object = 2.0 cm

Distance of object = 12 cm

Because object at focal point

We need to calculate the image distance of diverging lens

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-12}

v=\infty

The rays are parallel to the principle axis after passing from the diverging lens.

We need to calculate the image distance of converging lens

Now, object distance is ∞

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{34}-\dfrac{1}{\infty}

v=34

The image distance is 34 cm right to the converging lens.

Hence, The final image relative to the converging lens is 34 cm.

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Answer:

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b) 158.4kg-m^2/s

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Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation 

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a) From I = MK^2

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7 0
3 years ago
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