(29%) Problem 4: In the figure below, the converging lens'...

(29%) Problem 4: In the figure below, the converging lens' focal length is 14.5 cm, d = 14 cm, and the diverging lens' focal length is -13.7 cm. A 1.2 cm tall object is 7.9 cm to the left of the converging lens. d 25% Part (a) Where is the image formed, in cm, by the converging lens, measured from that lens (using negative for a virtual image)? Grade = 100% Correct Answer Student Final Submission Feedback Not available until end date s'_c = - 17.36 Correct! s'_c = -17.36 Grade Summary Deduction for Final Submission 0% Deductions for Incorrect Submissions, Hints and Feedback [?] 0% Student Grade = 100 - 0 - 0 = 100% Submission History All Date times are displayed in Eastern Standard Time Red submission date times indicate late work. Date Time Answer Hints Feedback 1 Sep 28, 2020 11:19 AM s'_c = - 17.36 s'_c = -17.36 25% Part (b) How tall, in cm, is the image formed by the converging lens (using negative for an inverted image)? 25% Part (c) Where is the image formed by the diverging lens, in cm, measured from that lens (using negative for a virtual image)? 25% Part (d) How tall, in cm, is the image formed by the diverging lens (using negative for an inverted image)?

Transcript

0:00 Hello students.

00:01 In this question we are provided with a converging and diverging lens and the values of different parameters have been provided to us.

00:07 So we have the focal length of the converging lens that is in positive and the focal length of the diverging lens.

00:13 So we'll take this to be f1 and takes to be f2.

00:16 Let d be the distance between the two lenses and h -i is the height of the h -o is the height of the object that is given to be equal to 1 .2 centimeters.

00:27 And we have u1 which is the initial, obviously.

00:30 Distance to be equal to 7 .9 taking sign convention this becomes minus 7 .9 centimeters because the distance is being calculated towards the left all distances calculated towards the right are considered to be positive and f1 that is the focal length of the converging lens is taken to be positive because for a convex lens we know that the focus lies on towards the right hand side whereas for the diverging lens we know that the focus lies on the left -hand side of the principal axis now we'll start with the various parts of the question in part one we are asked to find out what will be the image distance so we have the various values so we'll have 1 by f1 is equal to 1 by v1 minus 1 by u1 this implies that 1 by v1 is equal to 1 by f1 plus 1 by u1 and putting the values of the various parameters so we have 1 by 14 .5 and this is minus of and this is plus so we'll have minus of 1 point 1 by 7 .9 centimeters or we have v1 which is the image distance to be equal to 14 .5 multiplied with 7 .9 whole divided by 7 .9 minus of 14 .5 centimeters on calculating this we get v1 is equal to 11 .15 .55 divided by 6 .6 and this hole is in minus centimeters or or we get that v1 is equal to minus of 17 .35 centimeters.

02:06 So this is the image distance for the converging lens.

02:11 Now we'll find out what will be the height of the image form due to the converging lens.

02:16 We know that magnification is equal to the image distance by the object distance.

02:21 So we have m is equal to minus of 17 .35 divided by minus of 7 .9.

02:31 This gives to be equal to 2 .19 but we also know that magnification is equal to height of the image by height of the object this implies that height of the image is equal to height of the object multiplied with magnification and height of the object is given to be equal to 1 .2 and if we multiply it with the magnification due to converging lens that is 2 .19 this we get to be equal to 2 .628 2 .628 centimeters.

03:01 So this is the value of the value of the image of the 2.

03:03 Of the height of the image formed due to the converging lens.

03:08 Now we'll move on to the third part of the question in which it is asked to find out what will be the image distance due to the diverging lens...

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