An object 2.50 cm high is placed 12.0 cm in front of a thin lens of focal length 3.0 cm. Calc

An object 2.50 cm high is placed 12.0 cm in front of a...

An object $2.50 \mathrm{~cm}$ high is placed $12.0 \mathrm{~cm}$ in front of a thin lens of focal length $3.0 \mathrm{~cm}$. Calculate ( $a$ ) the image distance, $(b)$ the magnification, and $(c)$ the nature of the image.
(d) Check your answers by a graph.

Key Concepts

Ray Diagrams

Ray diagrams are graphical representations used in optics to trace the paths of light rays as they pass through a lens. They provide a visual confirmation of the calculated image distance, magnification, and overall image properties. By drawing key rays (such as the parallel ray, the focal ray, and the central ray), one can validate the application of the thin lens equation and the resulting image characteristics.

Image Formation and Nature

The nature of the image, including whether it is real or virtual, and inverted or upright, is determined by the lens and object configuration. Typically, for a converging lens with the object placed beyond the focal length, a real and inverted image is formed. Understanding these characteristics is crucial for predicting the qualitative aspects of image formation.

Magnification

Magnification in lens systems is defined as the ratio of the image height to the object height, and it can also be expressed as the ratio of the image distance to the object distance with a negative sign (m = -di/do). This concept is essential for determining how much larger or smaller the image will be compared to the object, as well as whether the image is inverted or upright.

Thin Lens Equation

The thin lens equation (1/f = 1/do + 1/di) relates the focal length, object distance, and image distance for a lens. It is fundamental in geometric optics as it allows for calculation of the position and properties of an image formed by a lens when the object distance and focal length are known.

Transcript

00:01 So in this question, we are given that there is a lens and in front of this lens an object is placed.

00:11 The height of the object is given to us.

00:14 It is 2 .5 centimeter and the distance of the object from the lens is also given to us, which is 12 centimeter.

00:24 We are also given the focal length of the lens, which is 3 centimeter.

00:29 So what we need to find is the image location.

00:34 This is the a part of the problem.

00:37 We have to find the image location.

00:39 In the b part, we have to find the magnification.

00:43 In the c part, we need to tell the nature of the image.

00:52 And in the d part, we need to support our answer with a diagram.

00:57 We need to draw definitely a ray diagram.

01:02 So let's start and solve this.

01:04 This is pretty simple we know for a lens 1 by v minus 1 by u is always equal to 1 by f now the data which is given to us we can write the data here and we can use that also we are given that the value of u is equal to minus 12 why minus 12 because this is my x -axis this is my y -axis so this distance is definitely negative so u is minus 12 focal length is plus 3.

01:40 So we can find the value of v very easily brown here.

01:45 Substituting these values here, what we are going to get 1 by v minus 1 by minus 12 is equal to 1 by 3.

01:58 If you solve this, you'll get the value of v to be equal to yes, 4 centimeter...

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