An object of height 3.0 cm is placed at 25 cm in front of a diverging lens of focal length 20 cm. Behind the diverging lens, there is a converging lens of focal length 20 cm. The distance between the lenses is 5.0 cm. Find the location and size of the final image.
di = _____ cm
hi = _____ cm
Height of image, "h=3.0 \\space cm"
Image distance, "u_1=-25\\space cm"
Focus, "f_1=-20\\space cm"
Let distance of image formed by the diverging lens "=v_1"
"\\dfrac{1}{f_1}=\\dfrac{1}{v_1}-\\dfrac{1}{u_1}"
"\\dfrac{1}{v_1}=-\\dfrac{1}{20}-\\dfrac{1}{25}"
"v_1=-\\dfrac{100}{9}\\space cm"
Height of image, "h_1'=h\\times\\dfrac{v_1}{u_1}"
"h_1'=\\dfrac{4}{3}\\space cm"
The image formed by the diverging lens will be the object for converging lens
Distance of object for converging lens, "u_2=-(v_1+5)"
"u_2=-\\dfrac{135}{9}\\space cm"
Focus, "f_2=20\\space cm"
Image distance "=v_2"
"\\dfrac{1}{f_2}=\\dfrac{1}{v_2}-\\dfrac{1}{u_2}"
"\\dfrac{1}{v_2}=\\dfrac{1}{20}-\\dfrac{9}{135}"
"v_2=-60\\space cm"
Height of the image formed, "h_2'=h_1'\\times\\dfrac{v_2}{u_2}"
"h_2'=\\dfrac{16}{3}=5.33 \\space cm"
Therefore distance of final image, di = 60 cm (left to the converging lens)
Height of final image, hi = 5.33 cm
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