Q1- an object is places 5 cm away from a concave lens that...

Q1- an object is places 5 cm away from a concave lens that has a focal length of 3 cm. use the lens equation to find the image location orientation, magnification and if its real or virtual. Q2- an object is places 5 cm away from a convex lens that has a focal length of 3 cm. use the lens equation to find the image location orientation, magnification and if its real or virtual. Q3- lens A to be a convex lens with a focal length of 20 cm, lens B is concave lens with focal length of 15 cm a- Calculate and compare the optical powers for lenses A, and B b- Find the combined optical power if the two lenses were cascaded behind each other Q4- Use ray tracing or Lens equation to find the image of an object 1 cm high located at 2 cm away from a convex lens that have a focal length of 4 cm. find the image type( real/virtual ) location, orientation, and magnification.

Transcript

00:01 In this question, we have a concave lens, right? now imagine we have a concave lens, something like this.

00:06 And we have some, this is the principle axis of this concave lens, and you have an object which is placed some problem here, and the focus of this, concave lens, something like this.

00:18 So you ask you to find the image, right? so suppose the image is somewhere a distance, d .i.

00:27 Away from the length according to the i then according to the length equation is 1 over di plus 1 over 5 centimeter must be equal to minus 3 centimeter right because this is the contab length so the focal lens you have to put a minus sign right so from this you can work out actually that 1 over di is given by you know 1 over 5 minus actually minus 1 for minus 1 over 3 right uh centimeter centimeter.

01:01 And this you get and actually is 15 centimeter, right? and you have three plus five eight.

01:10 So you find that the i is just given by minus 15 over eight centimeter.

01:17 And i use this is an active.

01:18 So it means that the image is actually virtual.

01:22 And it's of course a virtual image and the orientation is upright, right? it's upright, and the magnification is given by d .i over d .o.

01:37 That is given by minus 15 over 8 divided by 5, right? so it's actually a small image.

01:45 And the second one is very similar.

01:47 In the second case, you have your convex lens, right? in this case, the equation is given by 1 over d .i plus 1 over 5 centimeter equals 1 over 3 centimeter right so from this you'll find that and again the is the image the distance of image from the lens right so di is just given by 1 over 3 over 1 minus 5 over 1 and 1 zan meter actually and this is you'll find this is 15 divided by you know i think it's 2 right centimeter and actually now you see this image actually is it's a real image and it's a real image and it's actually upside down.

02:35 It's a real image and upside down.

02:38 And you can work out the magnification, which again given by this distance divided by five.

02:43 So it's again a smaller image.

02:47 And then there's cross -the -question.

02:50 You have a convex lens with focal lens of this.

02:54 And lens b is concave lens with focal lens of this...

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