Problem 40 not 39. Combination of Lenses I. A 1.20-cm-tall object is 50.0 cm to

Question

Problem 40 not 39.

Combination of Lenses I. A 1.20-cm-tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, (his one having a focal length of 60.0 cm. is located 300.0 cm to the right of the first lens along the same optic axis, (a) Find the location and height of the image (call it I_1) formed by the lens with a focal length of 40.0 cm. (b) I_1 is now the object for (he second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses. [(a) 100 cm to the right of the lens 1,4.80 cm tall, inverted; (b) 150 cm to the right of the second lens, 7.20 cm tall, upright] Combination of Lenses II. Repeat Problem 34.39 using the same lenses except for the following changes: (a) The second lens is a diverging lens having a focal length of magnitude 60.0 cm. (b) The first lens is a diverging lens having a focal length of magnitude 40.0 cm. (c) Both lenses are diverging lenses having focal lengths of the same magnitudes as in Problem 34.39. [(a) 262.5 cm to the right of lens 1,1.80 cm tall, inverted; (b) 73.7 cm to the right of second lens, 0.122 cm tall, inverted]

Explanation / Answer

Hi,

As per your request the only problem considered will be the 34.40.

In this case we have the following data:

h = 1.20 cm (height of the object)              p1 = 50.0 cm (distance of the object from the first lens)

f1 = 40 cm (focal length of the first lens)         f2 = 60 cm (focal length of the second lens)

d = 300 cm (distance between both lenses, the second one is at the rigth of the first one)

The equations to use in this problem are:

1/p + 1/q = 1/f ; where q is the distance of the image from the lens

M = - q/p = h' / h ; where M is the increase produced by the lens and h' is the height of the image.

We also have to remember certain things:

p is positive if the object is in front of the lens (to the left in this case)

q is positive if the object is behind the lens (to the right in this case)

f is positive if the lens is converging

h' is positive if the image is vertical

(a) If I understand correctly, then in this part the second lens is a diverging one, while the first is a converging one.

If so, then we have the following:

1/q1 = 1/f1 - 1/p1 = 1/40 - 1/50 = 1/200 :::::::: q1 = 200 cm

M1 = - q1/p1 = - 200/50 = - 4

h1' = h1M1 = (1.2 cm) (-4) = - 4.8 cm (this means that the image is inverted and it is bigger than the original object)

The value of p2 can be calculated with the distance between the lenses and the value of q1:

p2 = d - q1 = 300 cm - 200 cm = 100 cm

Then, the image of the first lens is used as the object of the second one:

1/q2 = 1/f2 - 1/p2 = 1/(-60) - 1/100 = -2/75 :::::::::: q2 = -37.5 cm

M2 = -q/p = - (-37.5)/100 = 0.375

h2' = h2M2 = (-4.8 cm)(0.375) = -1.8 cm (this means that the image is inverted)

The distance from lens 1 can be expressed as follows:

d' = d + q2 = 300 cm - 37.5 cm = 262.5 cm

(b) This part can be solved the same way than the previous one, but with the difference that the diverging lens is the first one.

1/q1 = 1/f1 - 1/p1 = 1/(-40) - 1/50 = -9/200 :::::: q1 = -22.2 cm

M1 = -q1/p1 = - (-22.2)/(50) = 0.444

h1' = h1M1 = (1.2 cm)(0.444) = 0.533 cm

p2 = d - q1 = 300 cm - (-22.2 cm) = 322.2 cm

1/q2 = 1/f2 - 1/p2 = 1/60 - 1/322.2 = 0.0136 :::::::: q2 = 73.5 cm

M2 = -q2/p2 = - 73.5/322.2 = - 0.228

h2' = h2'M2 = (0.533 cm)(-0.228) = - 0.122 cm (this means that the final image is inverted)

The distance to the second lens can be calculated as:

d' = q2 = 73.5 cm

(c) In this case, both lenses are diverging:

1/q1 = 1/f1 - 1/p1 = 1/(-40) - 1/50 = -9/200 :::::: q1 = -22.2 cm

M1 = -q1/p1 = - (-22.2)/(50) = 0.444

h1' = h1M1 = (1.2 cm)(0.444) = 0.533 cm

p2 = d - q1 = 300 cm - (-22.2 cm) = 322.2 cm

1/q2 = 1/f2 - 1/p2 = 1/(-60) - 1/322.2 = -0.0198 ::::::::: q2 = -50.5 cm

M2 = -q2/p2 = - (-50.5)/322.2 = 0.157

h2' = h2M2 = (0.533 cm)(0.157) = 0.084 cm (this means that the image is vertical)

The distance from the second lens is 50.5 cm to the left.

I hope it helps.

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