Two identical concave mirrors are set facing each other 1.0 m apart. A small lightbulb is placed

Two identical concave mirrors are set facing each other 1.0...

Two identical concave mirrors are set facing each other 1.0 $\mathrm{m}$ apart. A small lightbulb is placed halfway between the mirrors A small piece of paper placed just to the left of the bulb prevents light from the bulb from directly shining on the left mirror, but light reflected from the right mirror still reaches the left mirror. A good image of the bulb appears on the left side of the piece of paper. What is the focal length of the mirrors?

Key Concepts

Sign Conventions in Mirror Optics

Applying sign conventions correctly is crucial in mirror optics to determine the correct distances and image characteristics. Distances for objects and images may be positive or negative depending on whether they are real or virtual, and whether they lie in front of or behind the mirror. Using these conventions consistently helps in setting up and solving the mirror equation, especially in systems involving multiple reflections.

Mirror Equation

The mirror equation, 1/f = 1/do + 1/di, relates the focal length (f) of a mirror to the object distance (do) and the image distance (di). It is a fundamental tool for analyzing image formation by spherical mirrors, enabling one to predict the location, size, and nature of the image produced by a given mirror when an object is placed at a specific distance.

Real and Virtual Image Formation

In optical systems, an image can be real (formed by converging rays that actually meet) or virtual (formed by the apparent divergence of rays, which only seem to come from a location behind the mirror). Understanding the distinction between these two types of images is essential when analyzing systems where light is reflected more than once, as intermediate images can serve as objects for further reflections, often converting real images into virtual ones or vice?versa.

Virtual Objects in Multiple Reflection Systems

When a light ray is reflected by one mirror, the resulting image can act as a virtual object for another mirror. This occurs when the rays reaching the second mirror appear to diverge from a point behind it. This concept is key in systems with successive reflections, as the effective object distance for the second mirror must be calculated using proper sign conventions to account for the virtual nature of the intermediate image.

Transcript

00:01 As shown in this figure, there we have two mirrors, concave mirrors, let's say left mirror and this is right mirror, which are place facing each other and they are one meter apart from each other.

00:18 And in between them, we are placed here a source of light, a bulb exactly at the center of these two mirrors.

00:27 And in between this bulb and left mirror we are placed here a small piece of paper so that this bulb light will not shine off over to this left side concave mirror.

00:42 But still it is observed that this left mirror forms a very good image of this bulb.

00:51 And with this information we are supposed to find out what would be the focal length of this left.

00:58 Mirror.

01:00 Now, let us to solve this, to solve this, let us take the given data.

01:06 So, we have distance of bulb from a left mirror.

01:11 So let me call this as d distance of bulb from left mirror.

01:20 That is equal to 0 .5 meter.

01:23 Similar to this, the distance of bulb from a right side mirror that is also 0 .5 meter we have and the total distance between these two mirror that is given to us which is 1 meter.

01:41 Now by mirror equation we know that we have 1 upon that object distance plus 1 upon image distance is equal to 1 upon focal length of that mirror.

01:53 So this distance if i consider as the object distance then i can write this expression for image distance 1 upon v is equal to 1 upon f minus 1 upon this u where this u is same as that of distance which i consider for bulk from left or right mirror and let us simplify this so this we have we know that this is equal to we have 0 .5 so putting these values and simplifying i'll get this one upon v is equal to 0 .5 minus this f upon 0 .5 times f and taking inverse of this i'll get v is equal to 0 .5 f upon 0 .5 minus f.

02:43 Let's say this is our first equation...

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