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for violet light of wavelength 400 $\mathrm{nm}$ is $1.57 .$ A white object is placed 50.0 $\mathrm{cm}$ in front of this lens. Where will the red light and violet light be focused?for violet light of wavelength 400 $\mathrm{nm}$ is $1.57 .$ A white object is placed 50.0 $\mathrm{cm}$ in front of this lens. Where will the red light and violet light be focused?

$=71.1 \mathrm{cm}$The red image is focused at 64.9 $\mathrm{cm}$ and the violet image at 71.1 $\mathrm{cm}$ from the lens.

Physics 102 Electricity and Magnetism

Physics 103

Chapter 25

Optical Instruments

Electromagnetic Waves

Reflection and Refraction of Light

Cornell University

Hope College

University of Sheffield

University of Winnipeg

Lectures

02:51

In physics, wave optics is…

10:02

Interference is a phenomen…

09:56

. A thin planoconvex lens …

09:28

A thin double-convex lens …

05:11

If a diffraction grating p…

02:38

$\bullet$ If a diffraction…

04:29

A double-convex lens with …

02:32

A thin planoconvex lens ha…

02:50

A Lucite planoconvex lens …

04:31

(II) A diverging lens with…

02:19

Horizontal rays of red lig…

03:31

(II) A plan convex lens (…

02:28

$\bullet$ A converging len…

03:17

(a) Where does an object n…

02:25

Objects get their colors f…

01:08

White light is spread out …

02:23

The wavelengths of visible…

02:16

09:32

Monochromatic light with w…

00:26

A converging lens 7.20 $\m…

04:45

A narrow beam of white lig…

all right. So the formula to use here is one of her focal length equals Ah, next of infraction of lens minus one times one of her radius of curvature of the front and minus rate is one of Earth's curvature of the back end s o the sign convention for are one or two. Is that well for our one con cave in this positive convex is positive concave. It is negative for our two con cave. It is positive and convex. It is negative. So so double concave means that are one is minus 0.32 centimetres and our two is plus 20.24 centimeters on DH. So they're two parts to it. First is for red light. You have one of our aft equals one point 44 minus one times one over negative Teo point to be two centimeters plus one over 10.24 centimeters on DH. So you get that F is the focal length is negative. 19.2 centimeters. Sosa diverging lens with negative juggling. And so and so that a midge distance is one of her as prime equals one over F minus one over us. So that's, uh, remember, minus 19.22. Ah, hopes Talbot made a slight error. There. They'll be 31 negative. 31 point to pardon me, We'll be negative. 31.2. Um, plus one over. Playing five. Okay. And so you work that out and you get image distance of negative. 19.2 centimeters. Notice at it. There's a negative sign. No, in front of that. So images on the same side of violence. It is 19.2 centimeters in front of Linds. Same side of lenses. Object is a negative image. Distance. Ah, in a similar way, we're going to do violent light. I'm realizing that I should've done red with red ink. Well, violent light should be one over f equals one over a 5.7 minus one times one of her negative point to Tucson again. The same stuff there. Minus well, well over 0.24 centimeters. And so and so obstinate another area. There will be a negative over there. Um, and so what we get for for this one is that focal length is negative. 24.1 centimeters. Therefore, whatever image distances one over negative 24.1 Ah plus whenever 0.5 centimeters and and that gives us imaged Justin self negative. 16.3 centimeters, in other words, 16.3 centimeters in front of Lin's and that's it.

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