An image 3/2 of size of object is formed by a convex lens at a distance of 12cm from it. Find focal length

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Solution

h1(hight of object)=(2/3)h2(hight of image)
h1=(2/3)h2
but we know that,
h2/h1=-v/u
h1/h1(2/3)=-12/u
3/2=-12/u
u=-122/3
u=-42
u=-8cm

1/f(focal length)=1/v-1/u
by putting the value,
1/f=1/12-(-1/8)
=1/12+1/8
=2+3/24
=5/24
therefore,
f=24/5
f=4.8cm

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An image 3/2 of size of object is formed by a convex lens at a distance of 12cm from it. Find focal length

h1(hight of object)=(2/3)h2(hight of image) h1=(2/3)h2 but we know that, h2/h1=-v/u h1/h1(2/3)=-12/u 3/2=-12/u u=-12*2/3 u=-4*2 u=-8cm 1/f(focal length)=1/v-1/u by putting the value, 1/f=1/12-(-1/8) =1/12+1/8 =2+3/24 =5/24 therefore, f=24/5 f=4.8cm

h1(hight of object)=(2/3)h2(hight of image)
h1=(2/3)h2
but we know that,
h2/h1=-v/u
h1/h1(2/3)=-12/u
3/2=-12/u
u=-122/3
u=-42
u=-8cm

1/f(focal length)=1/v-1/u
by putting the value,
1/f=1/12-(-1/8)
=1/12+1/8
=2+3/24
=5/24
therefore,
f=24/5
f=4.8cm