# Combination of lenses. When two lenses are used in combination, the first one forms an image that...

## Question:

Combination of lenses. When two lenses are used in combination, the first one forms an image that then serves as the object for the second lens. The magnification of the combination is the ratio of the height of the final image to the height of the object. A 1.40 cm -tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm . A second converging lens, this one having a focal length of 60.0 cm , is located 300 cm to the right of the first lens along the same optic axis.

Part A Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0 cm . Enter your answer as two numbers separated with a comma.

Part B I1 is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

## Image Formation Through a Lens:

When an object is placed in front of a lens then an image of that object is formed due to the refraction of light through the lens.

The position of the image formed is related with object distance and focal length of the lens as

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

{eq}f {/eq} = focal length of the lens.

{eq}v {/eq} = image distance from the lens.

{eq}u {/eq} = object distance from the lens.

Magnification of the lens is defined as the ratio of height of the image formed by the lens {eq}I {/eq} to the height of the object {eq}O {/eq}.

{eq}m = \dfrac{I}{O} {/eq}.

Magnification of the lens is also defined as the ratio of image distance from the lens to the object distance from the lens.

{eq}m = \dfrac{v}{u} {/eq}.

## Answer and Explanation:

**Sign Convention:**

- All distances measured along the direction of incident ray of light are taken as positive and distances are taken as negative if...

See full answer below.

Become a Study.com member to unlock this answer! Create your account

View this answer

**Sign Convention:**

- All distances measured along the direction of incident ray of light are taken as positive and distances are taken as negative if measured along opposite the direction of incident ray of light.

**Assumptions and Given values:**

- {eq}O_1 {/eq} = height of object placed on left of first lens = {eq}1.40\ cm {/eq}.

- {eq}u_1 {/eq} = object distance from first lens = {eq}-50\ cm {/eq}.

- {eq}f_1 {/eq} = focal length of first lens = {eq}40.0\ cm {/eq}.

- {eq}f_2 {/eq} = focal length of second lens = {eq}60.0 \ cm {/eq}.

- {eq}d {/eq} = distance between first and second lens = {eq}300\ cm {/eq}.

- {eq}u_2 {/eq} = object distance from second lens.

- {eq}v_1,\ v_2 {/eq} = image distance from first and second lens respectively.

- {eq}m_1,\ m_2 {/eq} = magnification of first and second lens respectively.

- {eq}I_1 {/eq} = height of image formed by first lens.

- {eq}O_2 {/eq} = object height for lens 2.

- {eq}I_2 {/eq} = image height for lens 2.

**Part A:**

**Finding image distance for first lens.**

Using lens formula for the first lens.

{eq}\begin{align} \dfrac{1}{f_1}&=\dfrac{1}{v_1}-\dfrac{1}{u_1}\\ \dfrac{1}{v_1}&=\dfrac{1}{f_1}+\dfrac{1}{u_1}\\ &=\dfrac{1}{40}+\dfrac{1}{-50}\\ &=\dfrac{1}{40}-\dfrac{1}{50}\\ &=\dfrac{1}{200}\\ \Rightarrow v_1 &= 200\ cm. \end{align} {/eq}

**Finding image height for first lens.**

Magnification of first lens is given by

{eq}m_1 = \dfrac{v_1}{u_1}=\dfrac{I_1}{O_1}\\ \therefore I_1 = O_1\cdot \dfrac{v_1}{u_1}\\ =1.40\times \dfrac{200}{-50}\\ =-5.6\ cm. {/eq}

Negative sign indicates that image is inverted.

- {eq}v_1,I_1 = 200\ cm,\ -5.6\ cm {/eq}

**Part B:**

**Finding image distance for second lens.**

The image formed by first lens acts as object for second lens.

#### Learn more about this topic:

from UExcel Physics: Study Guide & Test Prep

Chapter 15 / Lesson 4