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## NCERT Class 12 Physics Textbook Chapter 9 With Answer PDF Free Download

### Chapter 9: Ray Optics and Optical Instrument

#### 9.1 INTRODUCTION

Nature has endowed the human eye (retina) with the sensitivity to detect electromagnetic waves within a small range of the electromagnetic spectrum.

Electromagnetic radiation belonging to this region of the spectrum (wavelength of about 400 nm to 750 nm) is called light.

It is mainly through light and the sense of vision that we know and interprets the world around us. There are two things that we can intuitively mention about light from common experience.

First, that it travels with enormous speed, and second, that it travels in a straight line. It took some time for people to realize that the speed of light is finite and measurable.

Its presently accepted value in a vacuum is c = 2.99792458 × 108 m s–1. For many purposes, it suffices to take c = 3 × 108 m s–1. The speed of light in a vacuum is the highest speed attainable in nature.

The intuitive notion that light travels in a straight line seem to contradict what we have learned in Chapter 8, that light is an electromagnetic wave of wavelength belonging to the visible part of the

spectrum.

How do reconcile the two facts? The answer is that the wavelength of light is very small compared to the size of ordinary objects that we encounter commonly (generally of the order of a few cm or larger).

In this situation, as you will learn in Chapter 10, a light wave can be considered to travel from one point to another, along with a straight line joining them.

The path is called a ray of light, and a bundle of such rays constitutes a beam of light. In this chapter, we consider the phenomena of reflection, refraction, and dispersion of light, using the ray picture of light.

Using the basic laws of reflection and refraction, we shall study the image formation by plane and spherical reflecting and refracting surfaces. We then go on to describe the construction and working of some important optical instruments, including the human eye.

### Reflection Of Light By Spherical Mirrors

We are familiar with the laws of reflection. The angle of reflection (i.e., the angle between the reflected ray and the normal to the reflecting surface or the mirror) equals the angle of incidence (angle between the incident ray and the normal).

Also that the incident ray, reflected ray, and the normal to the reflecting surface at the point of incidence lie in the same plane (Fig. 9.1).

These laws are valid at each point on any reflecting surface whether plane or curved. However, we shall restrict our discussion to the special case of curved surfaces, that is, spherical surfaces. The normal in this case is to be taken as normal to the tangent to the surface at the point of incidence.

That is, the normal is along the radius, the line joining the center of curvature of the mirror to the point of incidence.

### Refraction

When a beam of light encounters another transparent medium, a part of the light gets reflected back into the first medium while the rest enters the other. A ray of light represents a beam.

The direction of propagation of an obliquely incident ray of light that enters the other medium changes at the interface of the two media.

This phenomenon is called the refraction of light. Snell experimentally obtained the following laws of

refraction:

(i) The incident ray, the refracted ray, and the normal to the interface at the point of incidence, all lie in the same plane.

(ii) The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.

Remember that the angles of incidence (i ) and refraction (r ) are the angles that the incident and its refracted ray make with the normal, respectively.

Author | NCERT |

Language | English |

No. of Pages | 42 |

PDF Size | 710 KB |

Category | Physics |

Source/Credits | ncert.nic.in |

### NCERT Solutions Class 12 Physics Chapter 9 Ray Optics and Optical Instrument

*Question 1:*

*A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror with a radius of curvature of 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?*

*Answer*

Height of the candle, h =2.5 cm

Image size=h’

Object distance, u= -27 cm

The radius of the concave mirror, R= -36 cm

Focal length of the concave mirror, f=\frac{R}{2}=-18cm*f*=2*R*=−18*c**m*

Image distance=v\frac{1}{u}*u*1+ \frac{1}{v}*v*1= \frac{1}{f}*f*1\frac{1}{v}*v*1= \frac{1}{f}*f*1- \frac{1}{u}*u*1

=\frac{1}{-18}−181- \frac{1}{-27}−271=\frac{-3+2}{54}54−3+2=-\frac{1}{54}−541∴∴v=-54cm

Therefore, to obtain a sharp image, the distance between the screen and the mirror should be 54cm.

Image magnification is:m=\frac{h’}{h}=-\frac{v}{u}*m*=*h**h*’=−*u**v*∴ h’=\frac{-v}{u}\times h∴*h*’=*u*−*v*×*h*=-(\frac{-54}{-27})\times2.5=−(−27−54)×2.5=-5cm

5cm is the height of the image of the candle. As there is a negative sign, the image is inverted and virtual.

If the candle is moved closer to the mirror, then the screen needs to be moved away from the mirror so as to obtain the image.

*Question 2:*

*A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.*

*Answer :*

Size of the needle, h_{1}=4.5 cm

Object distance, u=-12 cm

The focal length of the convex mirror, f= 15 cm

Image distance= v\frac{1}{u}*u*1+ \frac{1}{v}*v*1= \frac{1}{f}*f*1\frac{1}{v}*v*1= \frac{1}{f}*f*1- \frac{1}{u}*u*1\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}=\frac{9}{60}151+121=604+5=609

∴ \frac{60}{9}960=6.7cm

The distance between the needle image and the mirror is 6.7cm and the image is obtained on the other side of the mirror.

Using magnification formula:m=\frac{h_{2}}{h_{1}}=-\frac{v}{u}*m*=*h*1*h*2=−*u**v*h_{2}=\frac{-v}{u}\times h_{1}*h*2=*u*−*v*×*h*1=\frac{-6.7}{-12}\times 4.5=−12−6.7×4.5=+2.5cmm=\frac{h_{2}}{h_{1}}=\frac{2.5}{4.5}*m*=*h*1*h*2=4.52.5=0.56

2.5cm is the height of the image. As it has a positive sign, the image is erect, virtual, and diminished.

As the needle is moved farther from the mirror, the image also moves away from the mirror resulting in a reduction in the size of the image.

*Question 3:*

*A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? Ifwater is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?*

*Answer:*

The actual depth of the needle in water, h_{1}=12.5cm

The apparent depth of the needle in water, h_{2} =9.4 cm

Refractive Index of water =\mu*μ*

The value of \mu*μ* can be obtained as follows:\mu=\frac{h_{1}}{h_{2}}*μ*=*h*2*h*1=\frac{12.5}{9.4}=9.412.5=1.33

Hence, 1.33 is the refractive index of water.

Now water is replaced by a liquid with a refractive index of 1.63

The actual depth of the needle remains the same, but its apparent depth changes.

Let x be the new apparent depth of the needle.\mu ‘=\frac{h_{1}}{x}*μ*‘=*x**h*1

Therefore y=\frac{h_{1}}{\mu ‘}*μ*‘*h*1=\frac{12.5}{1.63}=1.6312.5=7.67cm

The new apparent depth of the needle is 7.67cm. It is observed that the value is less than h_{2}*h*2, therefore the needle needs to be moved up to the focus again.

Distance to be moved to focus=9.4-7.67=1.73cm

NCERT Class 12 Physics Textbook Chapter 9 With Answer PDF Free Download