Refraction At Spherical Surfaces - Video Tutorials & Practice Problems

On a tight schedule?

Get a 10 bullets summary of the topic

1

concept

Refraction at Spherical Surfaces

Video duration:

5m

Play a video:

Video transcript

Hey guys in this video, we're gonna talk about refraction and image formation at spherical surfaces. We saw before when we talked about Snell's Law, how to apply refraction to a single ray of light across a flat boundary. Now we want to look at what happens when multiple rays of light come from an object refract through a spherical boundary and form an image inside of a different medium. OK. Let's get to it a single ray of light passing through a transparent surface undergoes refraction. All right, this is something we know we know how to apply Snell's law to figure this out. But many rays of light will also undergo refraction together an object placed in front of a surface, a transparent surface that allows transmission of light will form an image based on the focal length that that surface has. Those images can be real or virtual based on the shape of the surface. So that surface can have a positive focal length or it can have a negative focal length. All right, the image distance equation for a spherical surface is this where in one is the index of refraction that the object is in And in two is the index of refraction across the boundary that the light rays are passing into. OK. Now, there are some sign conventions that are important for this equation. First for a convex surface like the one in the image above me that I scrolled down past the radius is always considered positive. When you plug it into this equation for concave surfaces, the radius is considered negative. OK. And just like before, just like we had for mirrors, if you calculate a positive image distance, that is a real image that is inverted. And if you calculate a negative image distance, that's a virtual image that is upright. So this is exactly the same as it was for mirrors. OK. Let's do an example. An object in error is placed five centimeters in front of a transparent concave surface. If the radius of curvature is seven centimeters and the refractive index behind the surface is 1.44. Where is the image located is the image real or virtual? OK. Because this is a single surface refraction. We want to use our equation for that in one over the object distance plus N two over the image distance equals the difference between those indices divided by the radius. Now because this is a concave surface, the radius is going to be negative. Remember that's one of our rules and it's important to remember the sign convention. OK. What we have is our initial index or fraction which our initial medium is air is one. Our object distance we're told is five centimeters in front of the surface. So that's five centimeters. The index of a fraction behind the boundary is 1.44. Our image distance we don't know. And finally, our radius of curvature is negative seven centimeters. OK. It's important to remember that negative sign because if you don't, the answer is gonna be completely off. It's not just gonna be off by a sign. OK? So let's rearrange this equation to get N two over SI equals N two minus N one over R minus N one over. So, and let's plug in those numbers. This is gonna be 1.44 minus one over negative seven minus one over positive five, which if you plug into your calculator equals negative 0.263. OK. Let me minimize myself. Let me give myself just a little bit more space here. So we have N two over SI equals negative 0.263. If I multiply the si up and divide by negative 0.263 I get si is into over negative 0.263. The second refractive index is 1.44 and this answer is negative 5.5 centimeters. OK. Very simple to just apply the equation even though the arithmetic can get a little bit hairy. Now, is this image real or virtual? Don't forget the sign convention for images. If it's a negative image distance, it is virtual. The question didn't ask for it but the image is also upright because virtual images are always upright. All right guys, that wraps up our discussion on refraction at a single surface, a single spherical surface. Thanks for watching guys.

2

Problem

Problem

An object is embedded in glass as shown in the following figure. If the glass has a concave face, and is embedded in water, where will the image be located? Will the image be real or virtual?

A

-1.68 cm, virtual

B

-1.68 cm, real

C

-1.83 cm, virtual

D

-1.83 cm, real

E

No image is formed

Do you want more practice?

We have more practice problems on Refraction At Spherical Surfaces