An object 3.50 cm high is located 8.0 cm in front of a lens of -7.0 cm focal length. A lens o

step 1 So in this question, once again two lenses are given. So there are two lenses. Distance between

An object 3.50 cm high is located 8.0 cm in front of a...

Key Concepts

Thin Lens Equation

The thin lens formula relates the object distance, image distance, and the focal length of a lens using the formula 1/f = 1/do + 1/di. This relationship is fundamental in geometrical optics and is used to locate the position of an image for a given object when using a thin lens approximation.

Sign Conventions in Optics

Sign conventions determine the positive and negative values for distances and focal lengths in lens formulas. They are essential for correctly analyzing the image formation process. Generally, distances measured in the direction of the incoming light are positive, while those measured opposite are negative. Understanding these rules is critical when working with converging and diverging lenses.

Magnification

Magnification in lens systems quantifies the change in size of the image relative to the object and is given by the ratio of the image distance to the object distance (with a sign indicating image inversion). It provides valuable information on whether the image is enlarged or reduced and whether it is inverted or upright.

Compound Lens Systems

When more than one lens is used in sequence, the image formed by the first lens becomes the object for the next. Analyzing compound lens systems requires applying the thin lens equation sequentially. This concept is key to understanding complex optical instruments where multiple lenses are combined to achieve a desired overall magnification or image quality.

Ray Diagrams

Ray diagrams are visual tools used to represent the paths of light rays through lenses. They help in understanding how images are formed by tracing at least two rays from a point on the object. Drawing these diagrams to scale clarifies the spatial relations between object, lens, and image, and reinforces the understanding of lens behavior and image characteristics.

Transcript

00:01 So in this question, once again two lenses are given.

00:05 So there are two lenses.

00:10 Distance between the lenses is given to us, which is 3 .5 centimeters.

00:17 The vocal length of this one is minus 7 centimeter.

00:22 And the focal length of the second one is plus 4 .5 centimeter.

00:30 Then an object of some height is lying here.

00:35 This distance is given to us which is 8 cm.

00:41 The object height is given to us it is 3 .5 centimeters.

00:48 Now we have to find the location of the final image and also the size of the final image.

00:53 So we can apply lens formula.

00:57 We can say 1 by v minus 1 by u is equal to 1 by f, 1 by v.

01:04 I'm calling v1 the image formed by the first lens this one right minus one by u so u is minus 8 in my guess because it is lying here so that's why it is negative so minus 8 is equal to is equal to 1 by f so if it is minus 7 it is given so we can get the value of b1 from here the value of b1 if we calculate from here will come out to be minus 3 .73 so minus 3 .73 that means the image will be formed in this direction and this image i1 is at distance 3 .73 now this image will serve as an object for the second lens.

02:10 So we can see this is 3 .5 and this is from here to from here to here this whole distance would be 3 .73 plus 3 .5.

02:25 So that would be equal to 7 .23.

02:29 7 .23.

02:31 So 7 .23 distance will serve as an object for the second lens.

02:36 So i'm applying the lens formula for the second 1 by b minus 1 by u is equal to 1 by f 1 by i'm calling this b2 minus 1 upon u is negative here because definitely from this lens you need to go in this direction that's why it is negative so minus 7 .23 is equal to 1 by f f is positive which is 4 .5 you can see so we can get the value v2 on also from here, the value b2, we see it is coming out to 11 .92.

03:23 11 .92 centimeter...

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