By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where

q is the distance of the image from the lens

f = -10 cm is the focal length (negative for a diverging lens)

p = 15 cm is the distance of the object from the lens

Solving for q,

$\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}$

$q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm$

B) The image is upright

As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

$h_i = - h_o \frac{q}{p}$

where $h_i, h_o$ are the size of the image and of the object, respectively.

Since q < 0 and p > o, we have that $h_i >0$ , which means that the image is upright.

C) The image is virtual

As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.

This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual

4 0

3 years ago

What happens to a light wave when it travels from air into glass?

tino4ka555 [31]

Light is refracted when it crosses the interface from air to glass in which it moves more slowly.
Since the light speed changes at the interface, the wave length of the light must change too. The wave length decreases as the light enter the medium and the light wave changes direction.

7 0

3 years ago

Which part of an object's rate of change best defines acceleration?

grandymaker [24]

I believe it is speed.

Hope this helps!

5 0

3 years ago

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