# An object is 21.0 cm from a converging lens, and the image falls on a screen. When the object is...

## Question:

An object is 21.0 cm from a converging lens, and the image falls on a screen. When the object is moved 3.50 cm closer to the lens, the screen must be moved 2.20 cm farther away from the lens to register a sharp image. Determine the focal length of the lens.

## Spherical mirror:

There are two kinds of spherical mirrors: A concave mirror and a convex mirror. From the name itself, the objects is in the same side of the radius of the circle for the concave mirror, and if it the object is on the opposite side then it is a convex mirror. Spherical mirrors are very useful in magnifying object, a property which plane mirror doesn't have.

{eq}\dfrac 1f = \dfrac 1u + \dfrac 1v {/eq}

Where {eq}f {/eq} is the focal lenght of the mirror,

{eq}u {/eq} is the object distance, and

{eq}v {/eq} is the image distance.

## Answer and Explanation:

We can establish two different formula with the following given,

{eq}\dfrac 1f = \dfrac 1{21} + \dfrac 1v \\ \dfrac 1f = \dfrac 1{21-3.50} + \dfrac 1{v + 2.2} \\ {/eq}

Combining the two equation with {eq}f {/eq} as the common variable, we can solve for the image distance {eq}v {/eq}.

{eq}\dfrac 1{21} + \dfrac 1v = \dfrac 1{17.5} + \dfrac 1{v + 2.2} \\ \dfrac 1v - \dfrac 1{v + 2.2} = \dfrac 1{17.5} - \dfrac 1{21 }\\ \dfrac {2.2}{v*(v+2.2)} = \dfrac {1}{105} \\ v*(v+2.2) = 231 \\ v^2 + 2.2v - 231 = 0 \\ v = 14.14cm {/eq}

The focal length of the lens is

{eq}\dfrac 1f = \dfrac 1{21} + \dfrac 1{14.14} \\ f = \dfrac {14.14*21}{14.14+21} \\ f = 8.45cm {/eq}

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from Physics 101: Help and Review

Chapter 14 / Lesson 17