When two lenses are used in combination, the first one forms an image that then
ID: 1312036 • Letter: W
Question
When two lenses are used in combination, the first one forms an image that then serves as the object for the second lens. The magnification of the combination is the ratio of the height of the final image to the height of the object. A 1.40cm { m cm} -tall object is 59.0cm { m cm} to the left of a converging lens of focal length 40.0cm { m cm} . A second converging lens, this one having a focal length of 60.0cm { m cm} , is located 300cm { m cm} to the right of the first lens along the same optic axis.
s ? 1 , |y ? 1 | =Explanation / Answer
For the 1st lens,
Object height (ho) = 1.4 cm
Object distance (u) = -59 cm
Focal length (f) = +40 cm
(a) Using lens formula,
1/v - 1/u = 1/f
? 1/v = 1/u + 1/f
? v = uf/(u + f)
= (-59)(40)/(-59 + 40) = + 124.21 cm
? The image is formed at a distance of 124.21 cm to the right of the 1st lens
Magnifiction of 1st image = v/u = 124.21/(-59) = - 2.10
? Its magnified but inverted
? Height of image 1 = magnification x object height
= (-2.10) x 1.4 = - 2.94 cm
(b) Now, the 1st image serves as an object for the 2nd lens
As the separation between lenses is 300 cm and the 1st image is formed 124.21 cm to the right of the 1st lens,
? Object distance for 2nd lens (u) = -175.79 cm
Focal length of 2nd lens (f) = + 60 cm
? Fial image distance from 2 nd lens = v = uf(u+f)
? v_final = (-175.79)(60)/(-175.79+60) = +91.09 cm
Therefore, the final image is formed 91.09 cm to the right of the 2nd lens.
Magnification = v/u = (91.09)/(-175.79) = -0.52
Therefore, final image is reduced in size and inverted wrt the 1st image
? Final image height = magification x 1st image height
= (-0.52)(-2.94) = 1.53 cm
Thus, the final image is reduced in size by a factor of 1.53 and erect with respect to the 1st image
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