Question

DASED
An object is placed at a distance of 12cm in front of a concave mirror. It forms a real
image 4 times larger than the object. Calculate the distance of the image from the
mirror.
2. An object 5cm tall is placed on principal axis of a convex lens. Its 2cm tall image is
formed on the screen placed at a distance of 10cm from the lens. Calculate the focal
length of the lens.​

Answer 1

hey mates your answer is here

Object-size, h = + 5.0 cm; Object-distance, u = – 20.0 cm; Focal length, f = –15.0 cm; Image-distance, v = ? Image-size, h′ = ? Mirror formula is (1/v) + (1/u) = 1/f 1/v = (1/f) − (1/u) = (−1/15) + (1/20) = −1/60 v = − 60 cm. The screen should be placed at 60 cm from the mirror. The image is real. Magnification m = h'/h = − v/u = 60/(– 20) = – 3 Height of the image h' = mh = (– 3)x5 = − 15 cm The image in inverted and enlarged.

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Answer 2

question :-- (1)

m = - v/u

-4 = -v /(-12)

v = -48 cm

here m = -4 because the image is real ....

question :--- (2)

h₁=5cm

h₂=2cm

v=10cm

f = ????

m=h₁/h₂=v/u (Magnification)

we get,

u=vh₁/h₂

u = 10×5/2

u = 50/2

u = -25

now we will use lense formula....

1/v-1/u=1/f (Lens Formula)

=1/10-1/(-25)

LCM=250

=25+10/250

=250/35

f= 7.14cm

Hence the focal length is 7.14cm

this is your answer...

hope it helps to u ..