An object is placed 7.00 cm to the left of a converging lens of focal length 3.00 cm. A convex

An object is placed 7.00 cm to the left of a converging...

An object is placed $7.00 \mathrm{~cm}$ to the left of a converging lens of focal length $3.00 \mathrm{~cm}$. A convex mirror with a radius of curvature of $4.00 \mathrm{~cm}$ is placed $12.00 \mathrm{~cm}$ to the right of the lens. (a) Where is the intermediate image formed by the lens? (b) Is the intermediate image real or virtual and is it upright or inverted with respect to the object? (c) What is the magnification of this image? (d) Where is the image formed by the mirror? (e) Is the second image real or virtual and is it upright or inverted with respect to the original object? (f) What is the magnification due to the mirror? (g) What is the total magnification?

An object is placed $7.00 \mathrm{~cm}$ to the left of a converging lens of focal length $3.00 \mathrm{~cm}$. A convex mirror with a radius of curvature of $4.00 \mathrm{~cm}$ is placed $12.00 \mathrm{~cm}$ to the right of the lens. (a) Where is the intermediate image formed by the lens?
(b) Is the intermediate image real or virtual and is it upright or inverted with respect to the object? (c) What is the magnification of this image? (d) Where is the image formed by the mirror? (e) Is the second image real or virtual and is it upright or inverted with respect to the original object?
(f) What is the magnification due to the mirror? (g) What is the total magnification?

Key Concepts

Mirror Equation for Spherical Mirrors

Similar to the lens formula, the mirror equation (1/f = 1/d? + 1/d?) is used for spherical mirrors, such as convex mirrors. In this context, the radius of curvature relates to the focal length by f = R/2. This equation is vital for determining the properties of images formed by mirrors, accounting for the specific sign conventions related to convex (diverging) mirrors.

Magnification

Magnification describes how much larger or smaller an image is compared to the object, typically given by the ratio of image distance to object distance (with a negative sign for lenses to indicate image inversion). This concept helps in understanding the orientation of the image (upright or inverted) and is used to calculate the overall change in size in multi-element optical systems.

Sequential or Compound Optical Systems

When dealing with systems that include more than one optical element, such as a lens in combination with a mirror, the image formed by the first element often serves as the object for the subsequent element. Understanding how to sequentially apply the lens and mirror equations while tracking sign conventions is crucial for analyzing the overall system behavior and calculating total magnification.

Thin Lens Equation

This is the fundamental formula (1/f = 1/d? + 1/d?) used to relate the focal length of a lens with the object distance and the image distance. It is essential for determining where an image will form with respect to a converging lens and is applied with careful attention to the sign conventions for distances.

Sign Conventions in Optics

Optical systems require the correct application of sign conventions to properly identify whether distances are considered positive or negative. This determines if images are real or virtual and whether they are inverted or upright. It is crucial when applying both the lens formula and the mirror equation.

Transcript

00:01 In this problem, we have a conversion lens with this focal length, and a convex mirror with this radius.

00:12 And also we have an object that is in this location, seven certain mirrors in front of the lens, and the lens in the mirror are 12 centimeters apart.

00:26 In the first part of the problem, we want to compute where the image, produced by the length is.

00:36 So in order to compute this, we're going to use the lens equation.

00:42 So the lens equation for the lens is this, and we replace the values for the focal length, we have that this is 3.

01:02 The object distance is this value here for the length, so it's 1 over 7 plus 1 over q1, and from here we can compute the value of q1.

01:18 Q1 is 5 .25 centimeters.

01:28 So it's more or less at this point.

01:40 This is the location of the image produced by the first lens.

01:46 And in the second part of the problem, we want to know if this image that is producing this point here is a real, or a virtual image, but because we have compute the value of this q1 and this is positive, this means that this is a real image.

02:18 And in order to see if this is a inverted or upright image, when we want to compute the magnification.

02:30 So the magnification because of this lens is minus q1 over p1, this is minus 5 .25 celer meters over p1 that is 7 meters so this is just minus 0 .75 because this is minus it means that this image is inverted we could also say that this image is inverted because a converging lens produced an image that when this image is in the opposite side of the object is inverted.

03:22 But we can also see that from the sign of the magnification.

03:28 In the next part of the problem, we want to know where the image produced by the convex mirror is.

03:36 So for the convex mirror, the object is located at this point.

03:46 So for that, we can compute the object distance.

03:50 The object distance will be 12 minus 5 .25 and this is 6 .75 meters.

04:04 So this is the object distance.

04:08 And in order to compute the image distance, we're going to use the mirror equation.

04:22 So this is for the mirror.

04:26 So for the mirror, the focal length is a half of the radius...