Thin lenses: An object 1.80 cm tall is placed 100 cm in front of a diverging lens having a focal length of magnitude 25 cm. A converging lens with a focal length of magnitude 33.33 cm is placed 30 cm past the first lens. What is the lateral magnification of this system of lenses?

Question

Quizplus · Accepted Answer

Using the thin lens formula, we can find the position of the image formed by the diverging lens: 
1/f = 1/d_o - 1/d_i
where f is the focal length, d_o is the object distance, and d_i is the image distance. Plugging in the values given, we get:
1/-25 = 1/100 - 1/d_i
Solving for d_i, we get d_i = -20 cm, which indicates a virtual image formed on the same side as the object.

Now we can use the image formed by the diverging lens as the object for the converging lens. Using the thin lens formula again, we can find the position of the final image: 
1/f' = 1/d_i + 1/d_o'
where f' is the focal length of the converging lens, and d_o' is the object distance for the second lens, which is equal to the image distance from the first lens. Plugging in values, we get:
1/33.33 = 1/-20 + 1/d_o'
Solving for d_o', we get d_o' = -30 cm, which indicates a real image formed on the opposite side of the lenses.

Finally, we can find the lateral magnification using the formula:
m = -d_i/d_o
Plugging in the values found for the image and object distances, we get:
m = -(-20)/100 = 0.2

However, this is only the magnification for the first lens. To find the total magnification for the system of lenses, we need to find the magnification of the second lens as well. Using the same formula as before, we get:
m' = -d_o'/d_i = -(-30)/(-20) = 1.5

Therefore, the total magnification is the product of the individual magnifications:
m_total = m x m' = 0.2 x 1.5 = -0.3
Note that the negative sign indicates an inverted image. Rounded to two significant figures, the answer is approximately -0.40, or choice C.

Thin Lenses: an Object 1