An object is placed 18.2 cm from a first converging lens of focal length 14 cm.

Question

An object is placed 18.2 cm from a first converging lens of focal length 14 cm. A second converging lens with focal length 36 cm is placed 10 cm to the right of the first converging lens. (Take the direction to the right to be positive.)

(a) Find the position q1 of the image formed by the first converging lens.
cm

(b) How far from the second lens is the image of the first lens?
cm

(c) What is the value of p2, the object position for the second lens?
cm

(d) Find the position q2 of the image formed by the second lens.


(e) Calculate the magnification of the first lens.
M1 =

(f) Calculate the magnification of the second lens.
M2 =

(g) What is the total magnification for the system?
Mtotal =

(h) Is the final image real or virtual? Is it upright or inverted? virtual no image inverted real upright

Explanation / Answer

use the converging lens equation where 1/f=1/do+1/di and use the image of the first lens as image of the 2nd. i get a) 1/14=1/18.2+1/di and di=50.67 therefore virtual since the image is on the same side as the object (ie the positive side) b) 50.67-10=40.67 c) the object of 2nd lens is the image of the first so 40.67 d)1/36=1/40.67+1/di so di=313.516 so it is also virtual e) M=-di/do=hi/ho =-50.67/18.2=-2.78 so hi is negative so it's inverted f) M=-di/do=hi/ho =-40.67/313.516=-12750.7 so hi is a negative of a negative which is a positive so it's upright g) M=-di/do=hi/ho =-313.516/18.2=-17.2262 h) upright and virtual

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