An object 2.00 cm high is placed 44.0 cm to the left of a converging lens with a

Question

An object 2.00 cm high is placed 44.0 cm to the left of a converging lens with a focal length of 32.0 cm.>color="red">style="font-family: verdana, helvetica, sans-serif; font-size: 13px; line-height: 20px; ">color="red">style="font-family: verdana, helvetica, sans-serif; font-size: 13px; line-height: 20px; ">

A diverging lens with a focal length of -20.0 cm is placed 110 cm to the right of the converging lens.>color="red">style="font-family: verdana, helvetica, sans-serif; font-size: 13px; line-height: 20px; ">color="red">style="font-family: verdana, helvetica, sans-serif; font-size: 13px; line-height: 20px; ">

(Use the correct sign conventions for the following answers.)

An object 2.00 cm high is placed 44.0 cm to the left of a converging lens with a focal length of 32.0 cm. A diverging lens with a focal length of -20.0 cm is placed 110 cm to the right of the converging lens. (Use the correct sign conventions for the following answers.) Determine the final position of the final image.

Explanation / Answer

for converging lens

1/v1-1/u1=1/f1 where u1=-44 and f1=32 gives v1=117.33cms

for lens 2: 1/v2-1/u2=1/f2 where u2=-(110-117.33) and f=-20 cms

gives v2=11.57cms

that is 11.57 cms form the diverging lens towards right of it

the size of image was not asked in the question but if you wish to calculate you can calculate it as bject height * magnification where magnification=m1*m2 and for each lens m=-v/u final answer for

magnification=-4.21 negative sign suggests that it is inverted as compared to object

image height will be 8.42 cms invetred

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