An object 2.00 cm high is placed 40.0 cm to the left of a converging lens having a focal length

step 1 In the given problem we are having a combination of two lenses. The first one is a converging le

An object 2.00 cm high is placed 40.0 cm to the left of a...

An object $2.00 \mathrm{cm}$ high is placed $40.0 \mathrm{cm}$ to the left of a converging lens having a focal length of $30.0 \mathrm{cm} .$ A diverging lens with a focal length of $-20.0 \mathrm{cm}$ is placed $110 \mathrm{cm}$ to the right of the converging lens. Determine (a) the position and (b) the magnification of the final image. (c) Is the image upright or inverted? (d) What If? Repeat parts (a) through (c) for the case in which the second lens is a converging lens having a focal length of $20.0 \mathrm{cm}$.

An object $2.00 \mathrm{cm}$ high is placed $40.0 \mathrm{cm}$ to the left of a converging lens having a focal length of $30.0 \mathrm{cm} .$ A diverging lens with a focal length of $-20.0 \mathrm{cm}$ is placed $110 \mathrm{cm}$ to the right of the converging lens. Determine (a) the position and (b) the magnification of the final image. (c) Is the image upright or inverted? (d) What If? Repeat parts
(a) through (c) for the case in which the second lens is a converging lens having a focal length of $20.0 \mathrm{cm}$.

Key Concepts

Thin Lens Equation

The thin lens equation (1/f = 1/d_o + 1/d_i) is fundamental for understanding how a lens forms an image from an object. It relates the focal length of the lens to the object distance and the image distance, enabling the calculation of where an image will form under given conditions.

Sign Conventions

Sign conventions are essential for correctly applying the thin lens equation. They involve defining positive and negative distances for object positions, image positions, and focal lengths, which in turn allows for the correct interpretation of whether an image is real or virtual, and whether it is upright or inverted.

Magnification

Magnification describes the ratio of the image height to the object height and indicates the orientation of the image. It is calculated as m = -d_i/d_o, with the sign indicating whether the image is inverted or upright relative to the object, and its magnitude representing the degree of enlargement or reduction.

Sequential Image Formation in Composite Lens Systems

When dealing with systems involving more than one lens, the image produced by the first lens serves as the object for the second lens. Analyzing each lens sequentially using the thin lens equation and proper sign conventions is crucial for determining the overall image position, size, and orientation in composite optical systems.

Transcript

00:01 In the given problem we are having a combination of two lenses.

00:09 The first one is a converging lens of focal length f1 and the second one is a diverging lens of focal length f2.

00:32 The values of the focal lens are for f1.

00:39 This is plus 30 .0 centimeter and for f2 this is minus 20 .0 centimeter.

00:50 The height of the object and object has been kept to the left of this converging lens the height of this object at show has been given as 2 .00 centimeter and there is of this object from the converging lens has been given as p1 is equal to minus 40 .0 centimeter the magnitude of this distance is 40 centimeter and for sign convention we have taken the negative sign minus 40 centimeter now we have to obtain the final position of the image formed by this combination the gap between the two lenses has been given as 110 centimeter now first of all to find the position of the final image first we consider the converging lens so for converging lens the distance p1 is equal to minus 40 .0 centimeter the focal length f1 is plus 30 .0 centimeter so so using thin lens formula, 1 by q1 minus 1 by b1, where q is the distance of image from the lens is equal to 1 by f1.

02:30 So putting the known values, q1 is unknown, minus 1 by minus 40 is equal to 1 by 30.

02:38 So we get 1 by q1 is equal to 1 by 30 minus 1 by 40.

02:45 Whose lcm here will be 120 so this is 4 minus 3 means 1 by 120 so finally the image the distance of the image formed by this convex lens is the converging lens is 120 cm of positive means somewhere here at this point so this object this image this image for formed by the converging lens will serve as a virtual object for the diverging lens.

03:24 This image formed by converging lens will serve as a virtual object for diverging lens.

03:56 Hence for diverging lens the object distance p2 will be this remaining distance...

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