Question

A.

A 4.22 cm tall object is placed 6.80 cm in front of a convex mirror with focal length -2.65 cm.

How big does the image appear?

(include minus sign if inverted)

B.

A beam of light strikes a surface at 22 degrees from vertical. The beam proceeds through the new material at 49 degrees from vertical.

If the index of the original material is 1.50, what is the index of the new material?

Answer #1

**Part - A**

When the Object is lieing between infinity and pole of the
mirror, **the image is obatined between Pole and focus behind
the mirror. The obtained image is Virtual, Upright and
Dimished.**

**Part B**

From Snells law we know n1Sin1 = n2Sin2

n1 = 1.5, 1=22 , 2=49

Substituting in the equation we get

1.5 Sin22 = n2 sin49

n2 = (1.5 * sin22)/ sin 49 = 0.74452

**The refractive index of the new medium is
0.74452.**

A 28 cm tall object is placed in front of a concave mirror with
a radius of 37 cm. The distance of the object to the mirror is 94
cm.
Calculate the focal length of the mirror.
Calculate the image distance.
Calculate the magnification of the image (Remember, a negative
magnification corresponds to an inverted image).
Calculate the magnitude of the image height.

A 4.50 cm tall object is placed in front a convex mirror with a
focal length of -12.25 cm. If the magnification is 1/8, what is the
distance from the object to the mirror? Give your answer in
centimeters (cm) and with 3 significant figures.

A 31 cm tall object is placed in front of a concave mirror with
a radius of 32 cm. The distance of the object to the mirror is 88
cm.
Calculate the focal length of the mirror.
Tries 0/20
Calculate the image distance.
Tries 0/20
Calculate the magnification of the image (Remember, a negative
magnification corresponds to an inverted image).
Tries 0/20
Calculate the magnitude of the image height.

Consider a convex mirror with an object placed 20 cm away and a
virtual image 15 cm into the mirror. The image appears to be 2 cm
tall
a) What is the focal length of the mirror?
b) How large is the object?
c) Adding a convex convex lens at a distance of 10 cm from the
object (halfway between the object and the mirror), with a focal
length of 6 cm, where would the final image appear?
d) Removing...

object of height 25.0 cm is placed 50.0 cm in front of
a spherical mirror of focal length 35.0 cm. The image is formed on
the opposite side of the mirror. (2 points each)
a) Is the image real or virtual, and why?
b) Is the mirror concave or convex, and why?
c) Is the image upright or inverted, and why?
d) What is the image distance?
e) What is the image height?

An
object is located 20.2 cm in front of a convex mirror, the image
being 7.59 cm behind the mirror. A second object, twice as tall as
the first one, is placed in front of the mirror, but at a different
location. The image of this second object has the same height as
the other image. How far in front of the mirror is the second
object located?

An object is located 14.5 cm in front of a convex mirror, the
image being 3.73 cm behind the mirror. A second object, twice as
tall as the first one, is placed in front of the mirror, but at a
different location. The image of this second object has the same
height as the other image. How far in front of the mirror is the
second object located?

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height as the other image. How far in front of the mirror is the
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12.
An object 4.0 cmcm high is placed 17 cmcm in front of a convex
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4.1 cmcm
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-17 cmcm
24 cmcm
14. An object that is located at the center of curvature of a
spherical concave mirror produces which type of image?
A real, inverted image
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A virtual, inverted image
A virtual, upright image
There is no image for...

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