1. An object is placed 5.00 cm in front of a mirror whose...

1. An object is placed 5.00 cm in front of a mirror whose radius of curvature is 20.0 cm tall. If the object is 2.00 cm tall, a) Using two rays, draw the image diagram on the mirror diagram provided b) State the image characteristics c) Find the distance to the image and its height

Transcript

00:01 In this problem, we have been given that there is a concave mirror, and of course the focal length of this concave mirror will be negative.

00:09 So there is an object that's placed at a distance of 11 cm.

00:13 So we take the object distance as negative because it lies to the left of the pole.

00:19 And the mirror has radius of curvature 30 centimeters.

00:24 So we know that since this is a concave mirror, radius of curvature 2 will be negative.

00:28 And focal length here can be computed by just dividing this radius of curvature by 2, and this gets us the focal length as minus 15 centimeters.

00:38 And now we have to figure out the image distance.

00:42 So to compute the image distance, we will use the mirror formula.

00:46 That's 1 by u plus 1 by v is equal to 1 by f, and v here represents the image distance.

00:53 So putting the values here, we get minus 1 by 11 plus 1 by v is equal.

00:58 Equal to minus 1 by 15 and this gives us 1 by v as 1 by 11 minus 1 by 15 so simplifying this gives us 4 by 165 so from here we can figure out the image distance by reciprocating this 4 by 165 and that comes out to be approximately 41 .25 centimeters.

01:24 Also we observed that the image distance here that we figured out is positive.

01:29 That means the image lies to the right of this mirror.

01:34 And now we have to figure out the magnification.

01:36 So magnification is related to the image distance and the object distance using the formula m is equal to minus v by u.

01:43 So we just put the value here minus 41 .25 divided by u which was minus 11.

01:50 So when we divide 41 .25 with 11, we're going to get the magnification here as 3 .75.

01:58 And of course, this has no unit.

02:01 And now we have to figure out whether the image is real or virtual.

02:05 So we can see the magnification is positive, and that's only possible that the image and the objects are along the same orientation.

02:14 That means the image is upright, or we can see it is erect.

02:18 And of course, if the image is erect, then it will be virtual...

Question

1. An object is placed 5.00 cm in front of a mirror whose radius of curvature is 20.0 cm tall. If the object is 2.00 cm tall, a) Using two rays, draw the image diagram on the mirror diagram provided b) State the image characteristics c) Find the distance to the image and its height

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