Determine the image distance for an object d0 = 6.500 cm from a diverging lens o

Question

Determine the image distance for an object d0 = 6.500 cm from a diverging lens of radius of curvature 5.200 cm and index of refraction 1.800. (Express your answer as a positive quantity.)

Determine the image distance for an object do 6.500 cm from a diverging lens of radius of curvature 5.200 cm and index of refraction 1.800. (Express your answer as a positive quantity.) Number cm R=5.200 c' object What is the magnification of the object? (Be sure to insert the proper sign if needed.) Number What is the nature of the image? Inverted and smaller O Upright and smaller O Inverted and larger Upright and larger

Explanation / Answer

Focal length of the lens
1/f = (n-1)(1/R1 - 1/R2)
here the radus of both surface is same ,therefore
R1 = R2 = R , hence
1/f = (n-1)(-1/R - 1/R) = (1.8-1)(-2/5.2) = -0.307 cm
Now we know that the
1/f = 1/do + 1/di
where di is the image distance and do is the object distance
-0.307 = 1/6.5 + 1/di
di= - 2.169 cm
(b) Magnification = -v/u
= -(-2.169)/6.5 = 0.33
(c) Image will be upright and smaller.

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