An object is placed 10 cm to the left of a converging lens (Lens 1) with a focal

Question

An object is placed 10 cm to the left of a converging lens (Lens 1) with a focal length of 75 cm. Another converging lens (Lens 2) with a focal length of 4.1 cm is located 32.4 cm to the right of the first converging lens, as shown below. What is the location of the final image with respect to lens 27 Note that image distance can be positive (real image) or negative (virtual image). If your image distance is negative, put minus sign in front of your answer in the answer-box. Please round your answer to two decimal places. Lens 2 Lems Light of wavelength 616 nm is incident upon a single slit with width 3.6 x 10-4 m. The figure below shows the pattern observed on a screen positioned 217 cm from the slits. What is the distance, in cm. Please round your answer to two decimal places Light is incident on two slits that are separated by 179075 nm. The figure below shows the resulting interference pattern observed on a screen 854 mm from the slits. W wavelength, in nm, of light used in this experiment? Please round your answer to the nearest whole number. hat is the

Explanation / Answer

(a) Object distance from the lens 1 (do) = 10 cm
Focal length of the lens 1(f1) = 7.5 cm
Now using the lens equation , we know that
(1/do) +(1/di) = 1/f1
(1/10)+(1/di) = 1/7.5
di = 30 cm
where di is image distance , therfore the image formed by the first lens will be 30 cm right from the lens1.
Now this image will be works as object for the lens 2 , thefore the
object distance from the lens 2 (do2) = 32.4 -30 = 2.4 cm
Focal length of the lens 2(f2) = 4.1 cm
Now using the lens equation
(1/do2) +(1/di2) = 1/f2
(1/2.4)+(1/di) = 1/4.1
di2 = -5.79 cm
-ve sign indicate that final image is virtual .
hence the final image will formed at left of lens 2 .
di2 = -5.79 cm

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