Question

A 5.63 mm high firefly sits on the axis of, and 10.3 cm in front of, the thin lens A, whose focal length is 6.75 cm. Behind lens A there is another thin lens, lens B, with a focal length of 26.9 cm. The two lenses share a common axis and are 64.1 cm apart. Is the image of the firefly that lens B forms real or virtual? real virtual How far from lens B is this image located? Express the answer as a positive number. image distance from lens B: cm What is the height of this image? Express the answer as a positive number. image height: mm Is this image upright or inverted with respect to the firefly? upright inverted

Answer 1

* Solution - A B. 72563 12 10.3 cm 64 L4m 5,- 1 V u fA= 6.75cm fo= 26.9cm Gig.1) From lens formulai- where. fo=focal length of lensa ve distance of image from Given that, leng-A fA= 6.75cm u = distance of firefly. u from lens-A. L = ? then - 675 6.75 10.3cm V -10.3 to 6.75 10.3 69.525 6.75 X 10.3 10.3-6.75 #w(3.55) V=+19.58cm It means that the image of firefly will be form at distance 19.58 cm away from the lens-A. Now, this image will behave as an object for lens-B, and the distance of this new object from lens-B will be u'=(64, 1-19.58)

07, u = (64./- 19.58) 4'= 44.52 Since lens-B the distance is left side of the then u'- 44.52 Now we can calculate the distance of the formed by lens-B, and is given as- ㅗ ㅗ ㅗ where fe fo=focal length of lens-B Given, fo= 26.9 cm v'- image distance from lens-B. u'= -44.52 then - ㅗ ㅗ 26.9 v! -44.52 this abg to 44.52 v's 26.9 X 44.52 1197.588 44.52-26.9 17.62 v' = +67.97cm. It means that the final image of firefly will form at distance 67.97cm from lens-B.

by u Linear magnification of lens-A is given m = Y where y, y = height of Given that. . Che firefly) Y = 5:63mm y, = 5.63x102m. leng-A. 19.58cm. u= -10.3 cm the object Y = height of the image formed by V = then , - yz = vy 19.58 X5.63x102 4 -1013 42 = 110.2354x16? -10.7 x 10m -10,3 5-10.7 x 10 cm. Similarly, linear magnification of gens-13 ix given by where, Y u z = height of After putting the values of all above parameters we get a lens-B. Iz = 7 X Y 67.97X (-10.7x102) 727.279 X 10² =+16:34x102cm. 4 4.52 43=+0.16 am फू = 16:34 mm 1 Ang m2 = 7 the image formed by - 44.52 Y = or

Hence, the conclusions are given as ① The image of the firefly that being B forms will be- Real. 2) Image distance from lens-B= 67.9.7cm 3 Height of the image = 16.34 mm. 4 o the final image of the firefly will be Uprigut .