2. An object with size h 12cm is a distance s 15cm in front of a converging lens
ID: 1403158 • Letter: 2
Question
2. An object with size h 12cm is a distance s 15cm in front of a converging lens. The lens is L 18cm in front of a convex mirror. The lens has a focal length fi cm, and the mirror has a radius of R 48cm. fm (a) Considering just the lens, what is the value of s where the image is located (b) What is the height n' of the image? Is the image upright of inverted? Is the image virutal or real? (c) Now use the image of the lens as the object for the mirror. What is the object distance? (d) What is the location of the image from the mirror (e) What is the final height of the image? Is the image upright or inverted (relative to the original object)? Is the image virtual or real?Explanation / Answer
given,
focal length of the lens = 6 cm
distance of the object = 15 cm
by lens equation
1 / f = 1 / di + 1 / do
1 / 6 = 1 / di + 1 / 15
di = 10 cm
image is located 10 cm from the lens
height of image / height of object = -di / do
height of image / 12 = -10 / 15
height of image = -8 cm
since height is negative the image will be inverted
and the image position is another side of the lens so image is real so,
image is real and inverted
for convex mirror
object distance = 18 - 10 cm
object distance = 8 cm
focal length = -radius / 2
focal length = -48 / 2
focal length = -24 cm
by mirror equation
1 / f = 1 / di + 1 / do
-1 / 24 = 1 / di + 1 / 8
di = -6 cm
location of image is 6 cm from the mirror
height of image / height of object = -di / do
height of image / 8 = -(-6) / 8
height of image = 6 cm
relative to original object the image will be inverted because height is positive means upright image but the object is in inverted position so relative to original object the image will be inverted
image is virtual since image position is negative.
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