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 wanna 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. Okay, 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, a NA 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 really 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. Okay, 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 scroll down past the Radius is always considered positive. When you plug it into this equation Furqan cave surfaces, the radius is considered negative. Okay, and just like before, just like we had for mirrors. If you calculate a positive image distance, that is a really 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. Okay, let's do an example. A NA object in error is placed five centimeters in front of a transparent con cave 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 really or virtual okay, because this is a single surface refraction. We want to use our equation for that in one over the object distance plus in two over the image. Distance equals the difference between those indices divided by the radius. Now, because it is a con cave surface, the radius is going to be negative. Remember, that's one of our rules, and it's important to remember the sign convention. Okay, What we have is our initial index of refraction, 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 refraction behind the boundary is Our image distance we don't know. And finally, our radius of curvature is negative. Seven centimeters. 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. Okay, so let's rearrange this equation to get into over s I equals in two minus in one over R minus in one over S O. 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. Okay, minimize myself. Let me give myself just a little bit more space here. So we have in two over s I equals negative 0. If I multiply the S I up and divide by negative 0.263 I get s I is into over negative 0.263 The second refractive index is 1.44 and this answer is negative. 55 centimeters. Okay. Very simple to just apply the equation even though the arithmetic could get a little bit. Harry Now, is this image really 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 images also upright because virtual images air always upright. Alright, guys, that wraps up our discussion on refraction at a single surface. A single spherical surface. Thanks for watching guys
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?