An object 3.3 mm high is on the optical axis of two lenses with focal lengths f

Question

An object 3.3 mm high is on the optical axis of two lenses with focal lengths f1 = +5.0 cm, f2 = +10 cm. The object is 6.3 cm to the left of the first lens, and the second lens is 31 cm to the right of the first lens. Find the position (relative to the second lens) and the size of the final image by calculation. position size
An object 3.3 mm high is on the optical axis of two lenses with focal lengths f1 = +5.0 cm, f2 = +10 cm. The object is 6.3 cm to the left of the first lens, and the second lens is 31 cm to the right of the first lens. Find the position (relative to the second lens) and the size of the final image by calculation. position size f1 = +5.0 cm, f2 = +10 cm. The object is 6.3 cm to the left of the first lens, and the second lens is 31 cm to the right of the first lens. Find the position (relative to the second lens) and the size of the final image by calculation. position size position size

Explanation / Answer

1/p+ 1/q= 1/f 1/6.4 + 1/q= 1/5 solving for q you will get 22.86 cm. This distance is from the first lens . the image of the first lens will be used an object (virtual object) by the second lens. the distance of this virtual object from the second lens is (34-22.86) 11.14 cm. its magnification after the first lens is M1= -q/p= -22.86/6.4= -3.6 (inverted) final image distance from the second lens is : 1/11.14 + 1/q = 1/10 solving for q, you will get = 97.7 cm this is the distance of the image from the second lens. magnification by the second lens M2= -q2/p2= - (97.7/11.14) = -8.8 Total magnification, M= M1*M2= -3.6*8.8= 31.6 times magnified and with respect to the initial object. the final image size is h'= h*M = 31.6* 4.2mm= 132.6mm= 13.3 cm Therefore, Position: 98cm from the second size= 13.3 cm

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