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NCERT Exemplar Solution for CBSE Class 10 Science: Light- Reflection and Refraction (Part-III)

In this article you will get CBSE Class 10 Science chapter 10, Light- Reflection and Refraction: NCERT Exemplar Problems and Solutions (Part-III). Every question has been provided with a detailed explanation. All the questions given in this article are very important to prepare for CBSE Class 10 Board Exam 2017-2018.

May 31, 2017 11:33 IST
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Class 10 NCERT, Class 10 Science NCERT Problems and Solutions, Light Reflection and Refraction NCERT Exemplar ProblemsHere you get the CBSE Class 10 Science chapter 10, Light- Reflection and Refraction: NCERT Exemplar Problems and Solutions (Part-III). This part of the chapter includes solutions for Question No. 30 to 38 from the NCERT Exemplar Problems for Class 10 Science Chapter: Light- Reflection and Refraction. These questions include only the Long Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed explanation.

NCERT Exemplar Solution for CBSE Class 10 Science Chapter: Light‒Reflection and Refraction (Part-I)

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Science Board Examination 2017-2018 as well as other competitive exams.

NCERT Exemplar Solution for CBSE Class 10 Science Chapter: Light-Reflection and Refraction (Part-II)

Find below the NCERT Exemplar problems and their solutions for Class 10 Science Chapter, Light- Reflection and Refraction:

Long Answer Type Question

Question. 30 Draw ray diagrams showing the image formation by a concave mirror when an object is placed.

(a) Between pole and focus of the mirror

(b) Between focus and centre of curvature of the mirror

(c) At centre of curvature of the mirror

(d) A little beyond centre of curvature of the mirror

(e) At infinity

Answer.

(a) The image formation by a concave mirror when an object is placed between pole and focus of the mirror

image formation by concave mirror

(b) The image formation by a concave mirror when an object is placed between focus and centre of curvature of the mirror

 image formation by concave mirror

(c) The image formation by a concave mirror when an object is placed between focus and centre of curvature of the mirror at centre of curvature of the mirror

 image formation by concave mirror 

(d) The image formation by a concave mirror when an object is a little beyond centre of curvature of the mirror.

image formation by concave mirror

(e) The image formation by a concave mirror when the object is at infinity.

image formation by concave mirror

Question. 31 Draw ray diagrams showing the image formation by a convex lens when an object is placed

(a) Between optical centre and focus of the lens

(b) Between focus and twice the focal length of the lens

(c) At twice the focal length of the lens

(d) At infinity

(e) At the focus of the lens

Answer.

(a) The enlarged, virtual and erect image forms beyond 2F1 in the same side of object when the object is placed between optical centre and focus F1 of the lens.

image formation by convex lens

(b) The enlarged, real and inverted image forms beyond focus 2F2 on the other side of the object when the object is placed between focus F1 and twice the focal length of the lens.

image formation by convex lens

(c) The real and inverted image of equal to the size of object forms at focus 2F2 on the other side of the object when the object is placed at twice the focal length of the lens.

 image formation by convex lens

(d) The real, inverted and highly reduced image forms at focus F2 on the other side of the object when the object is placed at infinity.

 image formation by convex lens 

(e) The real, inverted and highly magnified image forms at infinity on the other side of the object when the object is placed at the focus of the lens.

image formation by convex lens

CBSE Class 10 Science Syllabus 2017-2018

Question. 32 Write laws of refraction. Explain the same with the help of ray diagram, when a ray of light passes through a rectangular glass slab.

Answer.

Laws of refraction of light are

  • The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
  • The ratio of sine of angle of incidence to the sine of angle of refraction is constant, for the light of a given wavelength and for the given pair of media. This law is also known as Snell’s law of refraction. The constant is also known as refractive index.

The ray diagram is as shown below:

 light passing through glass slab 

In the glass slab, the emergent rays are parallel to the incident ray because the extant of bending of the ray of light at the opposite parallel faced of rectangular glass slab are equal and opposite, so that emergent ray is parallel to incident ray.

Question. 33 Draw ray diagrams showing the image formation by a concave lens when an object is placed

(a) at the focus of the lens

(b) between focus and twice the focal length of the lens

(c) beyond twice the focal length of the lens

Answer.

(a) The image formation by a concave lens when an object is at the focus of the lens

 

image formed by concave lens  

(b) The image formation by a concave lens when an object is between focus and twice the focal length of the lens

image formation by concave lens

(c) The image formation by a concave lens when an object is beyond twice the focal length of the lens

image formation by concave lens

Question. 34 Draw ray diagrams showing the image formation by a convex mirror when an object is placed

(a) at infinity

(b) at finite distance from the mirror

Answer.

(a) The ray diagram is as follows

 image formation by convex mirror

(b) The ray diagram is given below

 image formed by convex mirror

Question. 35 The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80 cm and the lens?

Answer.

calculation of object distance by lens formula

Question. 36 Size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3rd to its size. At what distance, the object has been placed from the mirror? What is the nature of the image and the mirror?

Answer.

Here, it is not given whether the lens is concave of convex. So, we will do calculations for both types of mirror.

 

For concave mirror:

Focal length, f = ‒ 20 cm

Magnification, m = −1/3

Since, magnification, m = − v/u

Magnification, m = −1/3 = − v/u

⟹                   v = u/3                        

light reflection and refraction numericle problems

Hence, object must be placed at a distance 40 cm from the pole of convex mirror to form erect , virtual, and diminished image.

Question. 37 Define power of a lens. What is its unit? One student uses a lens of focal length 50 cm and another of ‒ 50 cm. What is the nature of the length and its power used by each of them?

Answer.

Power of lens is defined as the ability of a lens to bend the light rays. It is given by the reciprocal of focal length of the lens in meter. Its unit is dioptre.

Focal length = f = 50 cm, then P = 100/f = 100/ 50 = 2D, lens is convex.

If focal length = f = ‒ 50 cm, then P = 100/f = 100/ −50 = −2D, lens is concave.

Question. 38 A student focused the image of a candle flume on a white screen using a convex lens. He noted down the position of the candle, screen and the lens as under

Position of candle = 12.0 cm

Position of convex tens = 50.0 cm

Position of the screen = 88.0 cm

(i) What is the focal length of the convex tens?

(ii) Where will the image be formed, if he shifts the candle towards the lens at a position of 31.0 cm?

(iii) What will be the nature of the image formed, if he further shifts the candle towards the lens?

(iv) Draw a ray diagram to show the formation of the image in case as said above.

Answer.

Let f  be the focal length of the convex lens.

The distance of object should be measured from pole of the lens.

Distance of candle (or object) = Position of convex lens – Position of candle = 50 – 12 = 38 cm.

Now, by sign convention, distance of candle (or object) = u = – 38 cm

Similarly, distance of candle’s image = position of the screen – position of convex lens = 88 – 50 = 38 cm

By sign convention, distance of candle’s image = v = + 38 cm

lens formula numericle problems

The focal length of the convex lens is 19 cm.

(ii) When the candle is shifted towards the lens at a position of 31.0 cm.

Then, new object distance = position of convex lens c position of candle = 50 – 31 = 19

By sign convention, u = – 19 cm.

Now, focal length of the convex lens = 19 cm.  It means, the candle lays at the focus of lens, hence its image is formed at infinity.

(iii) When he further shifts the candle towards the lens. This means candle lies between optical centre and focus of convex lens, so, magnified, virtual and erect image of the candle will be formed.

(iv) The ray diagram of image formation is given below:

image formation by convex lens

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