MSBSHSE Solutions For SSC (Class 10) Science Part 1 Chapter 7- Lenses

MSBSHSE Solutions For SSC (Class 10) Science Part 1 Chapter 7 Lenses are available here for the students to practise thoroughly. Students can use these as it helps them to prepare most competently for their exam. We bring you Maharashtra Board Solutions for Class 10 designed by our subject experts. These solutions facilitate smooth and precise understanding of all the key concepts and topics that were covered in the chapter.

This chapter mainly focuses on the concept of Lenses and has detailed step-by-step solutions for the questions from the chapter. The chapter deals with Lenses, Sign convention, Defects of vision and their correction, use of lenses and other concepts related to the topic. Ray diagram for refracted light as well as Working of the human eye and lens are discussed at length in this chapter.

These solutions of MSBSHSE for Class 10 (SSC) have detailed explanations of the exercises that are found in the Maharashtra Board Science Textbooks for SSC Part 1. The Maharashtra State Board Solutions for Chapter 7 Science Part 1 is easily accessible here. Students can easily use it as a reference tool so that they can quickly revise all the topics for the exam. Students can prepare efficiently for the exams by studying with the help of these solutions. These solutions help them to study well and score high marks for the exam.

Maharashtra Board SSC (Class 10) Science Part 1 Chapter 7- BYJU’S Important Questions & Answers

1. Give a scientific reason why Simple microscope is used for watch repairs.

Answer: Simple microscope is a convex lens of small focal length and produces a virtual, erect and bigger image of the object. It is also called a magnifying lens as using such microscopes delivers a 20 times larger image of an object. Hence, they are used for watch repairs. Using it will ensure that the small parts and screws of the watch are visible.

2. 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. They are:

1. Refracting telescope, which uses lenses and

2. Reflecting telescope that uses mirrors as well as lenses

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.

3. Distinguish between Farsightedness and Nearsightedness.

Answer: There are different types of refraction defects. These include Farsightedness and Nearsightedness. In the case of nearsightedness, also known as myopia, the eye can see nearby objects clearly but distant objects appear indistinct. This is because the far point of the eye is closer to the eye instead of at the infinity. During this eye defect, the image of a distant object form in front of the retina. The defect is caused by two main reasons. They are: 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 eye defect can be rectified with the help of spectacles with concave lenses, which have proper focal length. These lenses diverge the incident rays and these diverged rays are then converged by the lens in the eye to form the image on the retina. Because the focal length of a concave lens is negative, a lens with negative power is required to correct nearsightedness. The power of the lens is different for different eyes depending on the magnitude of their nearsightedness.

Meanwhile, for farsightedness, you would be able to see distant objects clearly, but cannot see nearby objects that distinctly. This is because the near point of the eye is no longer at 25 cm but shifts farther away. During this defect the images of nearby objects get formed behind the retina.

The reasons for farsightedness are two:

1. The converging power of the lens becomes less as the curvature of the cornea and the eye lens decreases.

2. Due to the flattening of the eyeball the distance between the lens and retina decreases.

Now to correct this defect a convex lens with proper focal length has to be used. This lens converges the incident rays before they reach the lens and the lens, then converges them to form the image on the retina. The focal length of a convex lens is positive and for this reason the spectacles used to correct farsightedness has positive power. The power of these lenses is different depending on the extent of farsightedness.

4. 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: Here, the power of a lens is given as +1.5 D

Meanwhile, we know the power of lens = 1/focal length = 1/f

So, P =1/f

That is +1.5 D = 1/f

Hence, f = 1/1.5 = 10/15 = 2/3 = + 0.67 m.

So the focal length = + 0.67m.

Here, the lens is a converging type of lens or convex lens and the defect of the eye is Hyperopia or farsightedness.

5. 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: Assume that Object distance (u) = – 60 cm

Image Distance (v) = -20 cm

Apply the Lens formula,

1/ f = (1/v) – (1/u)

Replace with the values given,

1/f = 1/(-20)-1/(-60) = 1/60 – 1/20 = (1-3)/ 60 = – 2/60 = -30

Hence, f = -30 cm.

Thus the focal length of the lens = -30 cm.

Given that the focal length is negative, the lens is diverging.

6. Explain the centre of curvature and principal focus?

Answer: Centre of curvature of the lenses are the centres of spheres whose parts form surfaces of the lenses. A lens with both surfaces spherical, has two centres of curvature C1 and C2. Meanwhile, the principal focus of the lens is the point of the principal axis, where the light rays parallel to the principal axis converge, when they become incident on a convex lens.

7. What is the lens formula?

Answer: The lens formula shows the relation between the distance of the object (u), the distance of the image (v) and the focal length (f). It is given as 1/v- 1/u= 1/f. The lens formula is the same for any spherical lens and any distance of the object from the lens. It is however necessary to use the sign convention properly.

8. An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object?

Answer: Assume that

Height of the object (h1) is 5 cm,

focal length (f) is 10 cm,

distance of the object (u) is – 20 cm

Then calculate Image distance (v) = ?, Height of the image (h2) = ?,

and Magnification (M) = ?

Use the lens formula, 1/v-1/u=1/f

Then, 1/v=1/u + 1/f

1/v = 1/-20 + 1/10

1/v=(-1+2)/20 = 1/20

Therefore, v=20cm

The positive sign of the image distance shows that the image is formed at 20 cm on the other side of the lens.

Now, if M is magnification h2/h1= v/u

Then, h2= (v/u) × h1= 20/-20 × 5

h2= -1 × 5 = -5cm

And M = v/u = 20/-20=-1

Since, there is a negative sign to the height of the image and the magnification, it means that the image is inverted and real. It is below the principal axis and is about the same size as the object.

9. Draw a construction of the human eye.

Answer:

msbshse class 10 Science Part 1 Chapter 7 Question 9 Solution

10. What is the persistence of vision?

Answer: We are able to see an object because the eye lens creates its image on the retina. The image remains till the object is placed in front of him. However, it disappears when the object is taken away. This is not instantaneous and the image remains imprinted on our retina for 1/16th of a second, even after the object is removed. The sensation in the retina persists for a while and this is known as persistence of vision.

11. How do we recognize colours? Why are some people called colour blind?

Answer: Retina of our eye is made of light sensitive cells that are shaped like a rod and cones. These rod-like cells respond to the intensity of light and give information about the brightness or dimness of the object to the brain. It is the conical cells that respond to the colour and give information about the colour of the object to the brain. Brain processes these received information and we get to see the actual image of the object. While, the rod-like cells respond to faint light the conical cells do not. Thus, we perceive colours only in bright light. The cone cells can respond differently to red, green and blue colours. When red colour falls on the eyes, the cells responding to red light get excited more than those responding to other colours and we get the sensation of red colour. Some people lack the conical cells that respond to certain colours. These people are unable to recognize those colours or distinguish between different colours. These persons are said to be colour blind. Other than their failure to distinguish between different colours, their eyesight is normal.

Visit BYJU’S to get various guidelines and preparation tips for the exam and get into the merit list. Students can download the previous year papers to get an idea about the exam pattern. We also provide a ssc exam model question paper for the students.

 

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