A spherical mirror has a radius of -15.0 cm. An object 2.50 cm high is located in front of the

step 1 Given that we have a concave mirror. This is a concave mirror and this is the principal axis. An

A spherical mirror has a radius of -15.0 cm. An object 2.50...

Key Concepts

Magnification

Magnification in spherical mirrors is defined as the ratio of the image height to the object height and is also given by the formula m = -di/do. The negative sign indicates image inversion when the image is real. Magnification helps in understanding the size relationship between the original object and the image produced by the mirror.

Image Formation

Spherical mirrors form images through reflection, where the position and nature of the image depend on the object’s distance from the mirror. When an object is placed at different positions relative to the focal point, the reflected light can produce a real image (if the rays converge) or a virtual image (if the rays diverge), with real images being projected onto a screen and virtual images being observable only through the mirror.

Sign Conventions

In dealing with spherical mirrors, proper use of sign conventions is essential. The sign convention usually dictates that distances measured in the direction of the incident light are positive, while those measured against it are negative. For a concave mirror, typically the radius of curvature and the focal length are taken as negative values when using certain sign conventions, affecting the interpretation of the image distance and the nature (real or virtual) of the image formed.

Mirror Equation

The mirror equation relates the object distance (do), the image distance (di), and the focal length (f) by the formula 1/f = 1/do + 1/di. This equation is fundamental in the study of image formation by spherical mirrors, providing a method to calculate one of the parameters when the other two are known.

Transcript

00:01 Given that we have a concave mirror.

00:05 This is a concave mirror and this is the principal axis.

00:10 And there is an object lying on the principal axis.

00:13 The height of this object is given as 2 .5 centimeter.

00:20 And the distance of the object from the pole of the mirror is different in the different cases.

00:27 So we will assume it to be x and later on we can substitute the values.

00:31 And then what we have to find is we have to find the location of the image.

00:36 So this is a requirement.

00:38 So let us see how to find this.

00:39 This is very simple.

00:41 We know that 1 by v plus 1 by u is always equal to 1 by f.

00:49 So 1 by v will remain as it plus 1 by u would be minus of x.

00:57 Because the distance is in this direction.

00:59 So that's why we'll take it negative.

01:00 Okay and the focus is negative which is minus 15 it is given in the problem basically focal length is given as minus 15 centimeter this is the given data in the problem so which implies we can get the value of v from here uh that's pretty simple mathematics plus 1 by x so 1 by b is equal to this 15 x minus of x plus of 15 which implies v is equal to 15 x divided by 15 minus of x so this is the general value of the distance of the image now case wise we can solve it so in the case number one we are given the value of x to be equal to 45 centimeter so we can substitute that value here so we can see the value of v will be 15 into 45 divided by so we can see the value of v will be 15 into 45 by 15 minus 45 the answer will come out to be minus 22 .5 centimeter so the negative signs mean that if this is the mirror you have to go 25 centimeter in this direction these distances are taken as negative so this is here the final image will form so in the first case we have calculated the distance of the final image it is 22 centimeter to the left of the mirror in this case okay so now similarly we can find for case b also in the case b we are given the value of x to be equal to 30 centimeter we can substitute the value to get the value of image distance we know this is 15 into x which is 30 centimeter divided by 15 minus of x so this distance will come out to be minus 30 centimeter so again if this is our mirror, this is a pole, we have to go 30 centimeter in this direction to get the image distance.

03:18 So this would be your image.

03:20 So we have got the answer for the second part.

03:24 Now for third part which is part c, we are given the value of x to be equal to 15 cm...

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