A converging lens (focal length f1 ) and a diverging lens (focal length f2 = ?|

Question

A converging lens (focal length f1) and a diverging lens (focal length f2 = ?|f2 |) are arranged so that the lenses are a distance d from one another, as shown. If the distance d is greater than the sum of the two focal lengths but smaller than f1 (i.e., f1 + f2 < d < f1 with f2 < 0), then the two lens system acts as a converging lens with an effective focal length feff*. You can vary the focal length feff by varying the separation d. (In fact, this tunable focal length feff is an important feature of a camera zoom lens.) The two lenses shown have focal lengths f1 = 10.0 cm and f2 = -15.0 cm, but the distance d is unspecified.

note: *We can define the focal length of a compound lens as the image distance for incoming parallel lines (or an object at infinity). So, feff for our two-lens system is measured from the center of the converging lens.

(a) Find the image distance i1 for the initially parallel rays that pass through the diverging lens.

(b) Use your answer from part (a) to find p2 the object distance for the diverging lens. Hint: The image of the first lens is the object for the second lens and a virtual object has a negative object distance!

(c) Use your answer from part (b) to find feff the effective focal length of the two-lens system. Recall that feff is measured from the center of the converging lens.

(d) On the grid below, sketch a plot of feff vs. d.

note: For part D, just give me the values and i ll plot the graph on the grid myself. Thnkzz.

Explanation / Answer

F1 = focal length of converging lens

F2 = focal length of diverging lens

    F1 = 10 cm, f2 = - 15 cm

By lens formula , 1/f1 = 1/v1 – 1/u1                    

V1 = image distance of frst lens, u1 = object distance of frst lens.

1/f1 = 1/v1 – 1/ infinite                                 ( u1 = infinite)

1/10 = 1/v1 – 0

V1 = 10 cm

Now image distance of frst lens become object distance of second lens

U2 = object distance of diverging ( second lens) = -10 cm

1/f2 = 1/v2 – 1/u2

-1/15 = 1/v2 – ( - 1/10)

V2 = - 6 CM

THIS IS the ans of 1st part .image distance formed by diverging lens = 6 cm

(ii) object distance for diverging lens = u2 = 10 cm

(iii) since both the lenses act as converging lens.

So Feff = f1*f2 / f1+f2

           = 15*10 / 15+10

        = 6 cm

Now if two lenses ( convex, concave) are placed at a distance d

1/ Feff = 1/ f1 + 1/f2 – d/ f1*f2

1/6 = 1/10 +( 1/-15 )- d/ 10*-15

D = 20 cm

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