Maharashtra Board Class 10 Science Part – 1 Solution Chapter 7 – Lenses
Balbharati Maharashtra Board Class 10 Science Part – 1 Solution Chapter 7: Lenses. Marathi or English Medium Students of Class 10 get here Lenses full Exercise Solution.
|Maharashtra Class 10|
Science Part – 1 Solution
1) Match the columns in the following table and explain them.
|Column 1||Column 2||Column 3|
can be seen
object can be
|Near-sightedness||Problem on old age||Convex lens|
|Column 1||Column 2||Column 3|
|Farsightedness||Far away object can be seen clearly||Convex lens|
|Presbyopia||Problem on old age||Bifocal lens|
|Near-sightedness||Nearby object can be seen clearly||Concave lens|
2) Draw a figure explaining various terms related to a lens.
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.
4) Give scientific reasons:
a.) Simple microscope is used for watch repairs.
Ans: – A simple microscope can measure 20 times of magnification. If the object is placed in focal length of the lenses we see the effect and magnified image.
b.) can sense colours only in bright light.
Ans: – The retina present in our eyes helps to create image in our brain. The different cell present in our eyes helps to sense image. The rod cell helps to sense light in dim light whereas cone cell helps to sense the light in presence of light.
c) We can not clearly see an object kept at a distance less than 25 cm from the eye.
Ans: – The ciliary muscles present in our eyes maintain the focal length of our eyes by contracting and relaxing. The minimum focal length of our eyes is 25cm. So for this reason an object must be placed at 25cm from the eyes to see it clear.
5) Explain the working of an astronomical telescope using refraction of light.
Ans: – There are two lens in an astronomical telescope which are eyepiece and objective. As the focal length of the objective lens is more there are more light came to the object in parallel. This images which form by telescope are real, inverted and diminished.
6) Distinguish between:
a) Farsightedness and Near-sightedness.
b) Concave lens and Convex Lens.
Ans: – a) Farsightedness is the problem to see an object which is present near while the object is present in far away seen clearly. Whereas in short suitedness is a problem to see an object which is present in far away but which present in near seen clearly.
In farsightedness convex lense is used but in short-sightedness concave lens are used.
|Concave lens||Convex lens|
|Diverges the rays.||Converges the rays.|
|Image formed is a virtual and diminished.||Image formed is diminished as well as magnified.|
|Focus is virtual.||Focus is real.|
7) What is the function of iris and the muscles connected to the lens in human eye?
Ans: –Iris: – It control how much light should be entering in our eyes. By controlling the size of diaphragm muscles of pupil’s iris helps to maintain the eyes.
Muscles connected to the lens is a ciliary muscle. This ciliary muscles control the focal length by contracting and relaxing.
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?
Ans: – we know that the focal length of lens is f = 1/power.
Here power is +1.5D, so focal length f=1/1.5 = +0.67 cm.
As there is positive sign so the lens should be convex in nature.
And this problem is farsightedness.
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.
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?
Ans: – Let power (p1) be 2, p2 be 2.5 and p3 be 1.7D.
So total power is the summation of these power which is (2 + 2.5 + 1.7 =6.2) 6.2D.
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?
Ans: – According to lens formula, 1/v + 1/u = 1/ f.
Here given that u (object distance) =-60cm,v (image distance) =-20cm. F (focal length) =?
So, 1/f= 1/-60 + 1/-20 = 1/-30;
or, f=-30 cm.
The negative sign of focal length tells us that the lens is concave.
Here is your solution of Maharashtra Board Class 10 Science Part – 1 Chapter 7 Lenses
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