You plan to project an inverted image 30.0 cm to the right of an object. You have a diverging le

You plan to project an inverted image 30.0 cm to the right...

You plan to project an inverted image $30.0 \mathrm{cm}$ to the right of an object. You have a diverging lens with focal length $-4.00 \mathrm{cm}$ located $6.00 \mathrm{cm}$ to the right of the object. Once you put a second lens at $18.0 \mathrm{cm}$ to the right of the object, you obtain an image in the proper location. (a) What is the focal length of the second lens? (b) Is this lens converging or diverging? (c) What is the overall magnification? (d) If the object is $12.0 \mathrm{cm}$ high, what is the image height?

Key Concepts

Sign Conventions in Optics

Correctly applying sign conventions is critical in optics when using the thin lens equation and magnification formulas. Distances are measured with respect to the lens, and positive or negative values indicate the direction relative to a chosen coordinate system. Consistency in sign usage ensures that the calculated image positions and magnifications accurately reflect the physical behavior of the optical system.

Thin Lens Equation

The thin lens equation, 1/f = 1/d? + 1/d?, relates the focal length (f) of a lens to the object distance (d?) and the image distance (d?). This fundamental relation is used to determine where an image will form relative to a lens when an object is placed at a given distance, and it is essential for solving problems related to image formation in optical systems.

Magnification

Magnification in a lens system is defined as m = -d?/d?, where the negative sign indicates image inversion relative to the object. This concept explains how the size of an image compares to the size of the object and shows how multiple lenses in a system combine their magnification effects, which is important for determining the overall size and orientation of the final image.

Multi-Lens Systems

In systems that use more than one lens, the image formed by the first lens acts as the object for the next lens. Understanding the sequential nature of image formation in multi-lens (or compound) systems is crucial; one must apply the lens equation and magnification formula to each lens individually and then combine the results to determine the final image position, orientation, and size.

Converging vs Diverging Lenses

Converging (convex) and diverging (concave) lenses have fundamentally different optical properties, with converging lenses having positive focal lengths and diverging lenses having negative focal lengths. This difference affects how they form images; for instance, converging lenses can produce real, inverted images under the right conditions, while diverging lenses typically produce virtual, upright images. Recognizing these properties is key when selecting and combining different types of lenses.

Transcript

00:01 We have given initial condition of a mass of a gas and it is told that the gas exchange heat with surrounding.

00:09 First of all, in the first part of the question, it is told that it is slowly compressed to a volume of v0 by 2 and then suddenly compressed to a volume of v0 by 4.

00:23 And we have to find the final pressure.

00:24 So we have given that the value of cp by cv is equal to gamma and we have given initial pressure that is let that is p0 and initial temperature so initial volume is v0.

00:46 In the first part it is told that the gas is slowly compressed to a volume of v0 by 2.

00:55 So it is an isothermal compression.

01:03 So in isothermal compression we can see that p1, v1 is equal to p2 v2 here temperature is constant so now put the value of data so p1 is the initial pressure that is p .0 this is v0 is equal to p2 and the volume becomes v0 by 2 from here we get the pressure that is p2 is equal to 2 times of p .0 now further it is told that it is suddenly compressed to the volume v0 by 4 so now the process is adiabetic process because it is suddenly compressed so this is adiabatic process of compression so in this we can write that p1 v1 to the power gamma is equal to p2 v2 to the power gamma now put the value of p1 and v1 and all this data so this is we had calculated that is 2 p .0 volume becomes here v0 to the power gamma and pressure we have to find that is p2 and this is v0 by 4 to the power gamma.

02:22 From here we find the value of p2 that it comes to be p0 into 2 to the power gamma plus 1.

02:30 So this is the final pressure after isothermal compression and adiabatic compression.

02:38 So this is the answer for our first part...

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