A 5.33-mm-high firefly sits on the axis of, and 14.1 cm in front of, the thin le

Question

A 5.33-mm-high firefly sits on the axis of, and 14.1 cm in front of, the thin lens A, whose focal length is 5.33 cm. Behind lens A there is another thin lens, lens B, with focal length 30.1 cm. The two lenses share a common axis and are 58.3 cm apart. Is the image of the firefly that lens B forms real or virtual? How far from lens B is this image located (expressed as a positive number)? What is the height of this image (as a positive number)? Is this image upright or inverted with respect to the firefly?

Explanation / Answer

applying lens formula to the lens A: object distance u1= 14.1cm focal length f1= 5.33 cm applying the lens formula 1/f1 = 1/u1 +1/v1 => 1/5.33 =1/14.1 +1/v1 => 1/v1 = 1/5.33 -1/14.1 which gives us v1=8.56 cm from the lens. That means that this image acts as the object for the lens B. since the separation is 58.3 cm this image is 58.3 cm -8.56 cm from the second lens. soo u2= 49.74 and for lens B f2= 30.1cm that means 1/f2=1/u2+1/v2 => 1/30.1=1/49.74+1/v2 => 1/v2= 1/30.1-1/49.74 = .0131 which means that v2= 1/.0131 = 76.23 The image is real as the lens is convex lens and the object is beyond the focal point. the image is formed at 76.23 cm from the lens B. for the 1st lens height of the image= (V1/u1) * (height of the object) = (8.56/14.1) *5.33 = 3.23 cm this is the object for the second lens (lens B) height of final image = (v2/u2) * ( height of object for B)= (76.23/49.74)* 3.23 = 4.95cm the image is upright as the image formed by the 1st lens is inverted and that image is again inverted by the second lens to form the final image. So the final image is erect with respect to the firefly .

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