Question

Question 6 (1.5 points) This assignment is set up for sequential assessment. Complete each question and submit the answer before moving on to the next question. Correct answers for each question will be made available once the maximum number of attempts have been made for each submission. Lenses in Combination 2 Pay attention to the SIGN Convention. The sign of each quantity has meaning. Lenses in Combination 2 Divergiles Confringem Fi F F2 F2 As shown in the above figure, a diverging thin lens with focal points indicated by F1's is placed to the left of a converging thin lens with focal points indicated by F2's. The focal length of the diverging lens is -29.0 cm. The focal length of the converging lens is 20.0 cm. The distance between the two lenses is 59.0 cm. Case 1: An object is put 46.00 cm to the left of the diverging lens (Lens 1). The object is 14.0 cm tall. Find the image distance di 1 of the First image formed by the Diverging lens (Lens 1). Include a proper sign. Keep 2 decimal places. Enter a number cm Submit (5 attempts remaining) The height of the object is ho = 14.0 cm, what is the height of the First image hil? Include a proper sign, keep 2 decimal places. Enter a number cm Submit (5 attempts remaining) On which side of the Diverging lens (Lens 1) is the First image located? Select your answer from one of the following options. a. Left side b. Right side Submit (5 attempts remaining) Now make the transition from Lens 1 to Lens 2 (here Lens 2 is the Converging lens). The First Image is the Object (Object 2) to Lens 2. The focal length of the converging lens is 20.0 cm. The distance between the two lenses is 59.0 cm. Pay attention to on which side of Lens 2 is Object 2 located. This determines whether Object 2 is a REAL object or a VIRTUAL object.

Answer 1

1)For the diverging lens:

1/f = 1/v + 1/u

here, f = focal length of the diverging lens= -29 cm

v = image distance from the diverging lens.

u = object distance from the diverging lens= 46 cm

Now, 1/(-29) = 1/v + 1/46

=> 1/v = -(75/1334)

=> v = -17.79 cm

The distance of the image from the diverging lens= (-17.79) cm

We know,

hi/ho =- v/u

=> hi = -14×(-17.79/46)

=> hi = 5.41 cm

The height of the image is 5.41 cm

On the left side of the diverging lens the image will be formed.

2) The object distance for the converging lens= (59+17.79)= 76.79 cm

For the converging lens:

1/20 = 1/v + 1/(76.79)

=> 1/v = 0.0398

=> v = 27.04 cm

The distance of the image from the converging lens is 27.04 cm

The final image will be located on the right side of the converging lens.

The magnification= (-27.04/76.79)= -0.352

The final height of the image = (-0.352)×(5.41) = -1.905 cm

The final image will be real image.

Please comment if you have any doubt and like if it helps.

Happy learning.