A thin, converging lens having a focal length of magnitude 25.00 cm is placed 1.000 m from a p

A thin, converging lens having a focal length of magnitude...

Key Concepts

Sign Conventions in Optical Systems

Consistent sign conventions are critical in optics for determining whether distances and heights are positive or negative. Correct application of these conventions allows one to distinguish between real images (formed by actual convergence of light rays) and virtual images (formed by apparent divergence), and is essential for accurate use of formulas like the thin lens equation and magnification formula.

Combined Lens-Mirror Systems

In optical systems where a lens is combined with a plane mirror, the rays pass through the lens, reflect off the mirror, and then pass through the lens a second time. This double pass through the lens effectively alters the overall path and convergence of the rays, and the system can be analyzed by treating it as an equivalent optical system with a modified effective focal length.

Plane Mirror Imaging

A plane mirror creates an image that is the same size as the object and is located at the same distance behind the mirror as the object is in front of it. This principle is a key concept when analyzing optical systems that involve a mirror because it entails understanding how reflections can contribute to the formation of a final image when combined with additional optical elements like lenses.

Magnification and Image Orientation

Magnification in lens systems is defined as the ratio of the image height to the object height and is also given by the negative ratio of the image distance to the object distance (m = -di/do). This concept is important for determining not only the relative size of the image compared to the object but also whether the image is upright or inverted, based on the sign of the magnification.

Thin Lens Equation

This equation (1/f = 1/do + 1/di) is fundamental in geometrical optics and is used to relate the focal length of a thin lens to the distances of the object (do) and the image (di) from the lens. It is essential for calculating where an image will form when an object is placed at a given distance from the lens, and it helps in determining whether the image is real or virtual according to the sign conventions.

Transcript

00:01 For this problem on the topic of geometrical optics, we have a thin converging lens which has a focal length of 25 centimeters placed a meter from a plane mirror that is oriented perpendicular to the principal axis of the lens.

00:12 We have a flower then that is 8 .4 centimeters tall, a distance 1 .45 meters from the mirror along the principal axis of the lens.

00:21 And we want to know where the final image of the flower produced by the lens combination, lens mirror combination will be found.

00:29 We want to know if it's real or virtual, upright or inverted, and the height of the image.

00:36 Now, if the converging lens is replaced by diverging lens, having a focal length that is the same as the magnitude of the original lens, we want to recalculate the answers to part a.

00:48 Now, a thin converging lens with focal length, we'll call it faa, is equal to 25 centimeters, is placed a meter from a plane mirror on the same.

01:01 Principal axis the flower has a height h .o since it's the object of 8 .4 centimeters and is placed 1 .45 meters in front of the mirror which means that the object distance from the lens is 0 .45 meters which is 45 centimeters.

01:33 Now for part a for the image form formed by the lens.

01:40 Using the lens equation, we have one over the object distance d -o -a plus one over the image distance d -i -a is equal to one over the focal length of the lens f -a.

01:58 And so we can rearrange and solve for the image distance, d -i, which is f -a times d -a.

02:12 O over do minus f -a.

02:19 And so putting in our values, this is a focal length of 25 centimeters times the object distance, which is 45 centimeters, divided by 45 centimeters minus 25 centimeters.

02:42 And so calculating we get the image distance to be 56 .25 centimeters.

02:57 Now for the image formed by the mirror, the image formed by the lens is located 56 .25 centimeters behind the lens, which means the object distance for the mirror, we'll call it dlb, is equal to 100 centimeters minus 56 .24.

03:19 Centimeters, which is an object distance of 43 .75 centimeters.

03:28 So using the lens equation again, 1 over d -o -b plus 1 over the image distance, d -i -b gives us the focal length, 1 over the focal length of the mirror, fb.

03:48 So rearranging, we get 1 over the image distance.

03:56 D i b is equal to one over the focal length minus one over the object distance and putting in our values this is the focal length for the mirror is infinity so that's one over infinity minus one over the object distance so this means therefore that the image distance is equal to minus the object distance and this then is equal to minus 43 .8 centimeters.

04:45 And so the image that is formed by the mirror is located 43 .8 centimeters behind the mirror.

05:01 And the minus sign tells us that the image is virtual.

05:15 Now the total magnification, m total, is a total, is equal to the magnification due to the lens times the magnification due to the mirror m a times a b this is minus the image distance for the lens over the object distance for the lens for the lens multiplied by the same for the mirror minus d i over d o and so if we put in our values this is minus 56 .25 centimeters over 45 centimeters multiplied by minus 43 .8 centimeters divided by 43 .8 centimeters...

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