Class 10 Science Chapter 7 lenses Textbook Solutions

Maharashtra Board class 10 Chapter 7 lenses textbook Solutions Pdf 

Our Science Chapter 7 lenses class 10 questions and answers Pdf are highly effective for Maharashtra state board Exam. These Class 10 science lenses Maharashtra Board Solutions is extremely helpful when you are preparing for class 10 exams.

Solving all the questions of lenses chapter from the textbook without any trouble is a difficult task for class 10 science students. Most of the time Students get stuck with a particular Complex question. They try to attempt solving questions several times but are unable to find the correct answer. For these Our Class 10 Science textbooks Solutions have been studied & analyzed by ybstudys experts and the solutions have been created as per the Maharashtra state board syllabus and students understanding. Students can easily access these useful Solution of 10th Class Science lenses Class 10 Notes Pdf Download.

Cass 10 Science textbook Solutions will help a student to make a good score in their examination. So in order to help you with that, we have Compiled 10th Science Part 1 Chapter 7 exercise with answers for Class 10 below.


Class 10 lenses textbook solutions

Important Points to Remember about lenses

  1. Lenses are basically magnifying glasses with curved sides.
  2. A lens having both spherical surfaces puffed up outwards is called a convex lens or double convex lens or biconvex lens . 
  3. A lens having both spherical surfaces curved inwards is called a concave lens or double concave lens or biconcave lens 
  4. Cornea: The cornea is a thin and transparent cover (membrane) on the human eye through which light enters the eye. Maximum refraction of light rays entering the eye occurs at the cornea.
  5. Iris : The iris is a dark fleshy screen (muscular diaphragm) behind the cornea in the human eye. Its colours are different for different people.
  6. Pupil : The pupil is a small circular opening of changing diameter at the centre of the iris in the human eye
  7.  Uses of a convex lens: Simple microscope, compound microscope, telescope, camera, projector, spectrometer, spectacles, etc. 
  8. Uses of a concave lens : A concave lens is used in spectacles to correct myopia. It is also used in optical instruments 
  9. Use of a telescope : A telescope is used to observe a distant object such as mountain, moon, planet, star in the magnified form 
  10. FarsightednessIn this defect the human eye can see distant objects clearly but cannot see nearby objects distinctly NearsightednessIt's is also called as  Myopia In this case, the eye can see nearby objects clearly but the distant objects appear indistinct


Question 1. Match the following table and explain them. 

Column 1 Column 2 Column 3
Farsightedness Nearby object can be
seen clearly
Bifocal lens
Presbyopia Faraway object can be
seen clearly
Concave lens
Nearsightedness Problem of old age
Convex lens


Answer :
Column 1 Column 2 Column 3
Farsightedness Faraway object can be
seen clearly
Convex  lens
Presbyopia Problem of old age Bifocal  lens
Nearsightedness Near by object can be
seen clearly
Concave lens
 

Farsightedness : In this defect the human eye can see distant objects clearly but cannot see nearby objects distinctly. This means that the near point of the eye is no longer at 25 cm but shifts farther away. 
the images of nearby objects get formed behind the retina. 
There are two reasons for farsightedness. 
1. Curvature of the cornea and the eye lens 
decreases so that, the converging power of the lens becomes less. 
2. Due to the flattening of the eye ball the 
distance between the lens and retina decrease.
Farsightedness class 10 lens

There are two reasons for farsightedness. 
1. Curvature of the cornea and the eye lens 
decreases so that, the converging power of the lens becomes less. 
2. Due to the flattening of the eye ball the 
distance between the lens and retina decreases.
The focal length of a convex lens is positive thus the spectacles used to correct 
farsightedness has positive power. The power of these lenses is different depending on the extent of farsightedness.
farsightedness class 10 lenses

Presbyopia : Generally, the focusing power of the eye lens decreases with age. The muscles near the lens lose their ability to change the focal length of the lens. The near point of the lens shifts farther from the eye. Because of this old people cannot see nearby objects clearly. Sometimes people suffer from nearsightedness as well as farsightedness. In such a case bifocal lenses are required to correct the defect.

Nearsightedness : It's is also called as  Myopia In this case, the eye can see nearby objects clearly but the distant objects appear indistinct. This means that the far point of the eye is not at infinity but shifts closer to the eye.In nearsightedness, the image of a distant object forms in front of the retina 
Nearsightedness class 10 lenses

There are two reasons for this defect.
1.The curvature of the cornea and the eye 
lens increases.The muscles near the lens 
can not relax so that the converging 
power of the lens remains large.
2. The eyeball elongates so that the 
distance between the lens and the retina 
increases. 
This defect can be corrected by using spectacles with concave lens of proper focal length. The focal length of concave lens is 
negative, so a lens with negative power is required for correcting nearsightedness. .
Nearsightedness class10 lenses


Question 2. Draw a figure explaining various terms related to a lens.
Answer :
1. Centre of curvature (C) : The centres of spheres whose parts form surfaces of the lenses are called centres of curvatures of the lenses. A lens with both surfaces 
sperical, has two centres of curvature C1
 and C2
Center of curvature class 10 lenses

2. Radius of curvature (R) : The radii (R1
 and R2) of the spheres whose parts form surfaces of the lenses are called the radii of curvature of the lens.

3.Principal axis : The imaginary line passing through both centres of curvature is called the principal axis of the lens. 
Principal axis of the lens

4. Optical centre (O) : The point inside a lens on the principal axis, through which light rays pass without changing their path is called the optical centre of a lens. 
In  rays P1 ,Q1, P2, Q2 passing through O are going along a straight line. Thus O is the optical centre of the lens.
Optical axis class 10

5. Focal length : The distance between the optical centre and principal focus of a lens is called its focal length.
Focal length

Question 3. At which position will you keep an 
object in front of a convex lens so as 
to get a real image of the same size as 
the object ? Draw a figure.
Answer :  At 2F1


Class 10 science chapter 7 lenses


Question 4. Give scientific resons:
a. Simple microscope is used for 
watch repairs.
Answer : Simple microscope has convex lens which has the ability to produce 20 times larger as well as erect image of an object. This means the magnifying power of the microscope is very high. Thus, simple microscopes are used by watch makers to see the small parts and screws of the watch while repairing it.


b. One can sense colours only in 
bright light.
Answer : The cells present on the retina and responsible for colour vision are known as cone cells. These cells become active only under bright light and remain inactive under dark. Thus, we are able sense only in bright light.


c. We can not clearly see an object 
kept at a distance less than 25 cm 
from the eye.
Answer : When we try to see a nearby object, the eye lens becomes more rounded and its focal length decreases. Then a clear image of the object is formed on the retina of the eye. The focal length of the eye lens cannot be decreased beyond some limit. Therefore we cannot clearly see an object kept at a distance less than 25 cm from the eye.


Question 5. Explain the working of an 
astronomical telescope using 
refraction of light.
Answer : : In order to view distant objects clearly in their magnified form, the telescope is used and the telescopes used to see astronomical sources such as the stars and the planets are known as astronomical telescopes. 
There are two types of telescopes :
1. Refracting telescope, which uses lenses and
2. Reflecting telescope that uses mirrors as well as lenses
Working : In both the telescopes, the image formed by the objective is the object in the eyepiece, which produces the real image. Objective lenses are of large diameter and larger focal length so that the maximum amount of light coming from the distant object can be collected. However, the eyepiece is of smaller size with lesser focal length. Also, both the lenses fit inside a metallic tube in such a way that the distance between them can be changed. Meanwhile, the principal axes of both the lenses are on the same straight line. Generally, by using the same objective with different eyepieces, it is possible to get images with different magnification.

Explain the working of an  astronomical telescope using  refraction of light


Question 6. Distinguish between:
a. Farsightedness and Nearsightedness 
Answer :
Farsightedness Nearsightedness
It's is also called myopia It's is also called  hypermetropia
In these defect human eye is unable to
see nearby objects clearly
In these defect human eye  is unable to
see distant objects clearly
In these defect the image of a distant object is
formed in front of the retina
In these defect the image of a nearby object  
would be formed behind the retina
This defect can be corrected by using a concave lens This defect can be corrected
by using convex lens
 


b. Concave lens and Convex Lens 
Answer :
Concave lens Convex lens
Concave lens has its surfaces curved inwards Convex lens has its surfaces puffed up outwards
It's is thicker at the edges than in the middle It's is a thicker in the middle than at the edges
It's can form only a virtual image It's can form only a real image as well as virtual images
It's can form only a diminished image It's can be form a magnified diminished or the same sized image depending on the position of the object


Question 7. What is the function of iris and the muscles connected to the lens in 
human eye?
Answer : Function of Iris: The iris is a muscular diaphragm that controls the size of the pupil, which in turn controls the amount of light entering the eye. It also gives colour to the eye.

Function of ciliary muscles: The eye lens is held in position by the ciliary muscles. The focal length of the eye lens is adjusted by the expansion and contraction of the ciliary muscles.


Question 8: Solve the following examples.
i. Doctor has prescribed a lens having power +1.5 D. What will be the focal length of the lens? What is the type of the lens and what must be the defect of vision?
Answer : Given 
Power of lens, $P=+1.5 \mathrm{D}$ Now, focal length of lens, $f=\frac{1}{P}=+\frac{1}{1.5}=+0.67 \mathrm{~m}$
Since, the focal length is positive, the lens prescribed for correction is convex lens. Thus, the defect of vision is farsightedness or hypermetropia.


ii. 5 cm high object is placed at a distance of 25 cm from a converging lens of focal length of 10 cm. Determine the position, size and type of the image. 
Answer: Given
Height of object, $h_{0}=5 \mathrm{~cm}$ Object distance, $u=-25 \mathrm{~cm}$ Since the lens is converging, thus it is a convex lens. Focal length of the lens, $f=10 \mathrm{~cm}$ Using lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}=\frac{1}{10}+\frac{1}{-25}=\frac{3}{50}$
$\Rightarrow v=\frac{50}{3}=16.7 \mathrm{~cm}$
Thus, the image is formed $16.7 \mathrm{~cm}$ right of the lens.
Now, we know $\frac{v}{u}=\frac{h_{\mathrm{i}}}{h_{\mathrm{o}}}$
$\Rightarrow h_{\mathrm{i}}=\frac{50}{3 \times-25} \times 5=\frac{10}{3}=-3.3 \mathrm{~cm}$
The size of the image is 3.3 cm .negative sign shows that the image formed is real and inverted. Hence the image formed is real and inverted and diminished


iii. Three lenses having power 2, 2.5 and 1.7 D are kept touching in a row. What is the total power of the lens combination? 
Answer: Given
$P_{1}=2 \mathrm{D}, P_{2}=2.5 \mathrm{D}, P_{3}=1.7 \mathrm{D}$
Let the total power of the lens combination be $P$. Thus, $P=P_{1}+P_{2}+P_{3}=2+2.5+1.7$
$=6.6 \mathrm{D}$


iv. An object kept 60 cm from a lens gives a virtual image 20 cm in front of the lens. What is the focal length of the lens? Is it a converging lens or diverging lens?
Answer: Given
Object distance, $u=-60 \mathrm{~cm}$ Image distance, $v=-20 \mathrm{~cm}$ Using lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{f}=\frac{1}{-20}-\frac{1}{-60}=-\frac{1}{30}$
$\Rightarrow f=-30 \mathrm{~cm}$
Since, the focal length is negative, the lens is a diverging lens or a concave lens.


Maharashtra state board class 10 Science solutions part 1

Chapter 1 Science Part - I : Class 10 Science Chapter 1 - Gravitation Textbook Solutions 

Chapter 2 Science Part - I : Class 10 Science Chapter 2 - Periodic Classification of Elements textbook solutions.

Chapter 3 Science Part - I : Class 10 Science Chapter 3 - Chemical Reactions and Equations textbook solutions 

Chapter 4 Science Part - I : Class 10 Science Chapter 4 - Effects of electric current textbook solutions.

Chapter 5 Science Part - I : Class 10 Science Chapter 5 - Heat textbook solutions.

Chapter 6 Science Part - I : Class 10 Science Chapter 6 - Refraction of light textbook solutions.