Question

A converging lens of focal length 8.000 cm is 20.5 cm to the left of a diverging lens of focal length -6.00 cm. A coin is placed 12.5 cm to the left of the converging lens.

(a) Find the location of the coin's final image relative to the diverging lens. (Include the sign of each answer. Enter a negative value if the image is to the left of the diverging lens.) cm

(b) Find the magnification of the coin's final image.

Answer 1

The image of the first lens becomes the object for the second

So 1/do + 1/di = 1/f so 1/di + 1/17.8 = 1/7.93 ...so 1/di = 1/7.93 - 1/17.8 = 6.99x10^-2

Do di = 14.3cm..

So this becomes an object 17.8 - 14.3 = 3.5cm in front of the second lens

So 1/do + 1/di = 1/f2.......1/di = 1/f2-1/do = 1/(-6.08) - 1/3.5 = -0.450

So di = -2.22cm...which means it is 2.22cm in front of the second lens

Answer 2

The image of the first lens becomes the object of the second

So 1/s'1 + 1/s1 = 1/f.1.....=>1/s'1 = 1/f 1- 1/s1 = 1/8.04 - 1/11.4 = 0.03666

Therefore s'1 = 27.3cm

This means the object for the second lens is 27.3 - 24.3 = 3.0 cm behind the diverging lens...so s = -3.0 (a virtual object)

Now for the second lens we have 1/s'2 + 1/s2 = 1/f2

So 1/s'2 = 1/f 2- 1/s2 = 1/(-6.06) - 1/(-3.0) = 0.1683

So s'2 = 5.94cm

So the image forms 5.94 cm to the right of the diverging lens

Final magnification = m1*m2 = s'1/s1*s'2/s2 = 27.3/11.4*5.94/3 = 4.74

Answer 3

The image of the first lens becomes the object of the second

So 1/s'1 + 1/s1 = 1/f.1.....=>1/s'1 = 1/f 1- 1/s1 = 1/8. - 1/12.5 = 0.03666

Therefore s'1 = 27.3cm

This means the object for the second lens is 27.3 -20 =7.3cm behind the diverging lens...so s = -7.30 (a virtual object)

Now for the second lens we have 1/s'2 + 1/s2 = 1/f2

So 1/s'2 = 1/f 2- 1/s2 = 1/(-6.) - 1/(-7.3) =-0.029

So s'2 = -33.69cm

So the image forms -33.69 cm to the right of the diverging lens

Final magnification = m1*m2 = s'1/s1*s'2/s2 = 27.3/11.4*5.94/3 = 4.74

Answer 4

the final image will be formed infront of the deiverging lens 0.128 cm relativie to the diverging les ( infront of the diverging lens)

so its becoms