A diverging lens (f=-10cm) is located 20.0cm to the left of aconverging lens (f=

Question

A diverging lens (f=-10cm) is located 20.0cm to the left of aconverging lens (f=30.0 cm). A 3.00cm tall object stands to theleft of the diverging lens, exactly at its focal point.(A) Determine the distance of the final imagerelative to the converging lens. (B) What is theheight of the final image (including the proper algebraicsign)? I'm not sure where to begin. A diverging lens (f=-10cm) is located 20.0cm to the left of aconverging lens (f=30.0 cm). A 3.00cm tall object stands to theleft of the diverging lens, exactly at its focal point.(A) Determine the distance of the final imagerelative to the converging lens. (B) What is theheight of the final image (including the proper algebraicsign)? I'm not sure where to begin.

Explanation / Answer

with respect to diverging lens u=-10cm f=-10 so using 1/v-1/u=1/f we have 1/v=-1/10+1/-10=-2/20=-1/10 so v=-10cm this will be at a distance of 20+10 =30cm from the converginglens so with respect to converging lens u=-30cm f=30cm hence 1/v=1/f+1/u=1/30+1/-30 hence v= hence with respect to the converging lens the final image will beat infinity. B)when the final image is formed at infinity with respect toconverging lens the magnification=D/f=25/30 now the total magnification=-10/-10*25/30=5/6 now height of the final image=5/6*3=2.5cm

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