1) answer the question as posed in picture below, then: 2) if you swap the conve
ID: 1570830 • Letter: 1
Question
1) answer the question as posed in picture below, then: 2) if you swap the converging and diverging lens in the problem (converging lens at 25.0 cm from the object and the diverging lens 30.0 cm from the converging lens), where would you expect the image ? 3) if you put the converging lens 25.0 cm away from the object and you want the image to appear at 37.0 cm on the other side of the converging lens, where would you put the diverging lens ? Can you even get a real image that way ? 1) answer the question as posed in picture below, then: 2) if you swap the converging and diverging lens in the problem (converging lens at 25.0 cm from the object and the diverging lens 30.0 cm from the converging lens), where would you expect the image ? 3) if you put the converging lens 25.0 cm away from the object and you want the image to appear at 37.0 cm on the other side of the converging lens, where would you put the diverging lens ? Can you even get a real image that way ? 1) answer the question as posed in picture below, then: 2) if you swap the converging and diverging lens in the problem (converging lens at 25.0 cm from the object and the diverging lens 30.0 cm from the converging lens), where would you expect the image ? 3) if you put the converging lens 25.0 cm away from the object and you want the image to appear at 37.0 cm on the other side of the converging lens, where would you put the diverging lens ? Can you even get a real image that way ? 90. A small object is 25.0 cm from a diverging lens as shown in Fig. 23-68. A converging lens with a focal length of 12.0 cm is 30.0 cm to the right of the diverging lens. The two-lens system forms a real inverted image 17.0 cm to the right of the converging lens. What is the focal length of the diverging lens? 30.0 cm 17.0 cm FIGURE 23-68 H 25.0 cm Problem 90.Explanation / Answer
The image formed by the diverging lens will act as the object for the converging lens
For the converging lens,
(1/di) + (1/do) = 1/f
(1/di) + (1/17) = (1/12)
=> di = 40.8 cm
Therefore, the image formed by the diverging lens is 40.8 cm in front of the converging lens
Image distance for diverging lens = (30+25) - 40.8 = 14.2 cm
(1/di) + (1/do) = 1/f
=> (1/25) - (1/14.2) = (1/f)
=> f = -32.9 cm
Therefore, focal length of the diverging lens is 32.9 cm
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