An object with height 4.00 mm is placed 28.0 cm to the left of a converging lens that has foca

An object with height 4.00 mm is placed 28.0 cm to the...

Key Concepts

Sign Conventions

Proper use of sign conventions in optics determines the correct values for focal lengths, image distances, and object distances. They are crucial for correctly applying the thin lens equation and analyzing the orientations and positions of images produced in both single and multi-lens setups.

Thin Lens Equation

This principle relates the object distance, image distance, and focal length with the equation 1/f = 1/d_o + 1/d_i. It is essential for understanding how a single lens focuses light to form images and is a foundation for analyzing more complex multi-lens systems.

Magnification

Magnification is defined as the ratio of the image height to the object height and can also be expressed with the ratio of image distance to object distance (with a sign to indicate orientation). This concept helps determine how the size of an image is related to that of the object and whether the image is larger or smaller, as well as its orientation.

Image Orientation

Image orientation refers to whether an image is upright or inverted relative to the object. A negative magnification typically indicates that the image is inverted. Understanding image orientation is critical in problems where the final image’s inversion or non-inversion imposes conditions on the optical system.

Two-Lens Systems

In systems involving multiple lenses, the image formed by one lens serves as the object for the next. This sequential process requires careful tracking of object and image distances between lenses, as well as the propagation of magnification and sign conventions from one stage to the next.

Transcript

0:00 Hi.

00:02 In the given problem, height of the main object is h is equal to 4 .00 millimeter.

00:25 Distance of this object from the first lens, we can take it to be s1, and that is given as 28 .0 centimeter as it is kept towards the left side of the lens so we take it to be negative as per sign convention focal length of this convex lens f1 is given as 8 .40 centimeter and the gap between the two lenses is given as d is equal to 8 .00 centimeter.

01:20 Image distance formed by this first lens will be taken as s1 -dash object distance for the second lens will be taken as s2 and then image distance for the second lens will be taken as s2.

02:03 Dash.

02:05 The height of object is h, then height of image formed by first lens will be taken as h -dash, and height of image formed by the second lens will be taken as h -dableness.

02:36 And this h -dash will serve as the height of object for the second lens.

02:45 So height of object for the second lens will be h -dash only like this.

03:03 So in the first part of the problem, height of final image formed by the first lens is given as 5 .60 millimeter and that is a real image so it will be taken as negative because real images are always inverted so height of real image formed by the first lens and formed by the second lens height of the real image and that is the final image form by second lens and we will take it to be h double dash that is given as minus 5 .60 millimeter and negative sign has been used because the image is real so inverted so for the first lens using thin lens equation which is the relation between object distance image distance and the focal length of the lens and that is given as 1 by s dash minus 1 by s 1 by s 1 .1 is equal to 1 by f1 so plugging in the known values 1 by s 1 by minus 1 by minus 28 is equal to 1 by 8 .40 what we can say is 1 by s1 dash will come out to be 1 by 8 .4 minus 1 by 28.

05:22 So if we take here lcm 8 .4 into 28, so this is 28 minus 8 .4.

05:33 So if we solve this s1 dash, means the distance of image from the first lens comes out to be 8 .4 into 28 divided by 28 minus 8 .4 and it comes out to be 12 .0 centimeter.

05:56 Their positive sign sign signifies that this image is being formed towards the right side of this first lens.

06:06 Here now we can draw the image, the ray diagram.

06:12 This is the first lens which is a convex lens.

06:18 Another lens, we don't know the nature of this second lens.

06:23 So just draw a line for it.

06:26 The gap between them is given as 8 .00 centimeter.

06:34 Now, this is the focal length of this first lens means here this is f1.

06:43 And the object has been kept at a distance of 28 .0 centimeter.

06:54 Now, image is being formed due to this first lens at a distance of here, let it be i -dash.

07:03 So that distance we have just found to be 12 .0 centimeter.

07:10 This image formed by first lens will serve as virtual object for another lens.

07:17 So the distance of object, virtual object, from the second lens means s2, will come out to be 12 minus 8 means this is 4 .00 centimeter positive.

07:34 So we can write it here also.

07:38 Image.

07:41 We should write real image.

07:43 Real image formed by the first lens will behave like virtual object for the second lens...

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