A 4.0-cm tall object is positioned 12.0 cm in front of a converging lens. The fo

Question

A 4.0-cm tall object is positioned 12.0 cm in front of a converging lens. The focal length of the converging lens is 8.0 cm. 20 cm behind the converging lens is a diverging lens with a focal length of -12.0 cm. (This means the same point is used for the focus of each lens.) (i) Draw rays from the object to the converging lens that will form the first image from that lens. Then use that image as an object for the second lens and draw rays (not necessarily the same rays) that will go to the second lens and form a final image. (ii) Use the thin lens formula to check your answer for the position of the final image. (iii) What is the height of the final image?

Explanation / Answer

the 40 cm lens is placed at 10 cm to the left of the 20cm lens. If i place the object at 60 cm in front of the 40 cm lens I find using 1/f=1/p+1/q that q=+120cm , here f=40cm. p=60cm then q is positive, that means that the image is real and situated 120 cm to the right of the 40 cm lens. This real image becomes the object of the second lens, but here we have the perfect situation for a virtual object. In reality the image from the first lens is intercepted before it can form. . Soo for the second lens p= -110, f=20 and then q-+16.92 cm. This second image is real (+ sign) and situated at the right of the second lens. It is normal to find the final image inside focal lenght of the second lens. It's because the object is virtual, that means that rays of light were already converging when entering the second lens. the height is given by 2cm X - (120/60) X - (16.92/-110) the negative sign that apperars finally here means that the final image is inverted compared to the object. It is hard to draw the diagram of this problem because the virtual obect is difficult to visualise....

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