The left face of a biconvex lens has a radius of curvature of magnitude 12 and t

Question

The left face of a biconvex lens has a radius of curvature of magnitude 12 and the right face has a radius of curvature of magnitude 18 cm. The index of refraction of the glass is 1.44. Calculate the focal length of the lens for light incident from the left. One end of a long glass rod (n_glass = 1.5) is formed into a spherical surface with radius of curvature + 6 cm. An object 1 cm tall in air and along the axis of the rod, is located at 20 cm the vertex. Calculate the location and the size of the image this spherical surface forms of the object.

Explanation / Answer

(a) 1/f= (u-1)(1/r1 +1/r2)

1/f=(1.44-1)(1/12+1/18)

1/f=11/180

f=16.36 cm.

(b) we use lensMakers formula

n1/p + n2/q = n2-n1/R

1/20+1.5/q= (1.5-1)/6

1.5/q=0.5/6-1/20

1.5/q= 1/30

q/1.5=30

q= 45 cm.

so image distance= 45 cm

now magnification= m=-q/p= -45/20=-2.25

m= Hi/Ho

Hi= m Ho

Hi=-2.25 (1 cm)

Hi=-2.25 cm

so height of image= 2.25 cm .

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