Combination of Lenses I. A 1.20 -cm-tall object is 50.0 cm to the left of a converging lens of f

(a) First, we need to find the image formed by the first lens. We can use the lens formula, whic

Combination of Lenses I. A 1.20 -cm-tall object is 50.0 cm...

Combination of Lenses I. A 1.20 -cm-tall object is 50.0 $\mathrm{cm}$ to the left of a converging lens of focal length 40.0 $\mathrm{cm} . \mathrm{A}$ second converging lens, this one having a focal length of $60.0 \mathrm{cm},$ is located 300.0 $\mathrm{cm}$ to the right of the first lens along the same optic axis. (a) Find the location and height of the image (call it $I_{1} )$ formed by the lens with a focal length of 40.0 $\mathrm{cm} .$ (b) $I_{1}$ is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

Transcript

00:06 Here, the first lens is a convex lens of focal length 40 cm.

00:16 For convex lens, the focal length is positive.

00:20 Now, an object ab of height h1 equal to 1 .2 centimeter.

00:30 This is a positive.

00:32 It means it is erect.

00:38 So, ab is shown with arrow upward.

00:41 It is placed at a distance of 50 cm from the convex lens so u1 is equal to 50 centimeter the sign is negative because to measure the u we need to travel in the opposite direction of the incident rays now the ray diagram is as shown the ray which is traveling means parallel to the principal axis passes through the focus after reflection and the ray which is passing directly through the center of the means lens it passes undeflected and where the raise meet we have the image a dash and the image is formed at an distance of v1 so first let us find out the v1 so for the image formation by the this is under the heading image formation by lens f of focal length f one in this case u1 is equal to minus 50 centimeter because we are traveling opposite to the incident rate f1 is equal to 40 centimeter because for convex lens focal length is positive and h1 is equal to plus 1 .2 two centimeter it is an means it is an erect object and we know the lens formula is one upon v1 minus one upon u1 is equal to one upon f one it means one upon v1 is equal to one upon u1 plus one upon f1 now we put just the values 1 upon f1 is equal to minus 1 upon 50 and plus 1 upon 40 so if you saw this v1 comes out to be 200 centimeter the positive sign shows that v1 is on the right of the lens because to measure v1 we need we travel in the same direction as the direction of the raise right so v1 is positive now we calculate this is the location of the object now we calculate its magnification we know the magnification m1 by the first lens is equal to v1 upon u1 and is equal to h2 upon h1 now what is v1 v1 is 200 u1 is minus 50 it means magnification is minus four.

03:55 Magnification is minus four means h2 is equal to minus four times h1.

04:02 H1 is 1 .2.

04:05 So this is minus four times 1 .2.

04:09 It means minus 4 .8 centimeter.

04:15 H2 is minus 4 .8 centimeter...

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