Class 10 - Physics - Light Reflection Refraction

**Question1.**

Which one of the following materials cannot be used to make a lens?

(a) Water (b) Glass (c) Plastic (d) Clay

Answer:

Correct option ** (d) Clay**. Clay is an opaque material while water, plastic and glass are transparent. Therefore clay is not used to make a lens.

** Question2.**

The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?

(a) Between the principal focus and the centre of curvature

(b) At the centre of curvature

(c) Beyond the centre of curvature

(d) Between the pole of the mirror and its principal focus.

Answer:

Correct option (d) __between the pole of the mirror and its principal focus. __

The image formed by a concave mirror will be virtual, erect and magnified when the object is placed only between the pole and the principal focus of the mirror.

Object is between focus and the principal focus of the lens.

**Question 3.**

Where an object should be placed in front of a convex lens to get a real image of the size of the object?

(a) At the principal focus of the lens

(b) At twice the focal length

(c) At infinity

(d) Between the optical centre of the lens and its principal focus.

Answer:

Correct option ** (b) at twice the focal length**.

When the object is placed twice its focal length, an inverted, equal size and real image is formed by the convex lens.

**Question 4.**

A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be

(a) both concave.

(b) both convex.

(c) the mirror is concave and the lens is convex.

(d) the mirror is convex, but the lens is concave.

Answer:

Correct option __(a) both concave. __

According to sign convection focal length is always negative for concave mirror and concave lens.

**Question 5.**

No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

(a) plane.

(b) concave.

(c) convex.

(d) either plane or convex.

Answer:

Correct option ** (d) either plane or convex**.

A plane mirror and a convex lens always form a virtual and erect image.

**Question 6.**

Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

(a) A convex lens of focal length 50 cm.

(b) A concave lens of focal length 50 cm.

(c) A convex lens of focal length 5 cm.

(d) A concave lens of focal length 5 cm.

Answer:

Correct option (c) A convex lens of focal length 5cm.

When the object is between radius of curvature and focal length a magnified image is formed. So a convex lens with this focal length will produce magnified image.

**Question 7.**

We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm.

What should be the range of distance of the object from the mirror? What is the nature of the image?

Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

Answer:

Given:-

Focal length of the concave mirror= -15cm.

In order to get an erect image object must be placed between the pole of the mirror and its focus.

Image formed will be erect, virtual and enlarged.

**Question 8.**

Name the type of mirror used in the following situations.

(a) Headlights of a car.

(b) Side/rear-view mirror of a vehicle.

(c) Solar furnace.

Support your answer with reason.

Answer:

- Headlights of a car: - A concave mirror is used in the headlights of the car because a concave mirror can produce powerful parallel beam of light when the light source is placed at their principal focus.
- Side/rear-view mirror of a vehicle: - A convex mirror is placed side/rear view mirror of a vehicle.
- Convex mirror gives erect, virtual and diminished image of the object when object is placed in front of it.
- As a result we will have wide view. Therefore driver is able to see most of the traffic behind him.
- Solar furnace: - A concave mirror is used in solar furnace because concave mirror is a convergent mirror.
- They converges light incident on them at a single point known as principal focus. So sun’s rays can be concentrated at a single point to produce heat in solar furnace.

**Question 9.**

One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object?

Verify your answer experimentally. Explain your observations.

Answer:

Yes, it will. Even if half of a convex lens is covered with a black paper, the lens will produce a complete image.

But the intensity of the image may be less. It can be observed experimentally by using a lighted candle and a convex lens.

It produces a complete image of the object.

Experiment:-

- Take a lighted candle and place it in front of a convex lens mounted on a plane surface.
- Move the candle along the axis of the plane to obtain the full image on the screen. Once the full image is observed, mark the position of the candle without moving it.
- Now cover the lower exact half of the lens with a black opaque paper. Do not change the position of the candle.
- At this point, we will observe a full image of the candle, but we will find that the intensity is reduced.
- This is because the covered part of the lens does not allow light to pass through it. So the amount of light reaching the screen is reduced.

** Question 10.**

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm.

Draw the ray diagram and find the position, size and the nature of the image formed.

Answer:

Given:-

Size of object: 5cm

Distance of object (u): –25cm

Focal length (f) = 10cm

Distance of image = (v)

Using Lens Formula for converging lens,

= (1/ v) – (1/u) = (1/f)

=> (1/v) – (1/-25) = (1/10cm)

=> (1/v) = (1/10cm) + (1/-25cm)

=> (1/v) = (5-2)/ (50) cm

=> (1/v) = (3/50) cm

=> v= (50/3) =16.66cm

Magnification is given as,

m = (distance of image) (v)/ (distance of object) (u)

=> m = (16.66)/ (-25cm) = - 0.66

Now m = (height of image) (h’)/ (height of object) (h)

=> -0.66 = (h’)/ (5cm)

h' = -0.66 x 5cm

=>Therefore height of image (h’) = -3.3cm

(-ive) sign of height of the image shows an inverted image is formed.

Position of image: At 16.66cm on opposite side of lens.

Size of image: – 3.3 cm, i.e. smaller than object.

Nature of image: Inverted and real

** Question 11.**

A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Answer:

Given,

Focal length of concave lens (f) = – 15cm

Distance of image (v) = – 10cm

Distance of object = (u)

Using Lens Formula for converging lens,

= (1/ v) - (1/u) = (1/f)

=> (1/-10cm) – (1/u) = (1/f)

=> (1/u) = (-1/10cm) + (1/15cm)

=> (1/u) = (-3 +2)/ (30) cm = (-1/30) cm

=> u = -30cm.

Thus, object is place at distance of 30cm from the lens i.e. at 2F_{1} of the concave lens.

(-) ive sign shows the object is at 30cm in front of the lens.

**Question 12.**

An object is placed at a distance of 10 cm from a convex mirror of focal length

15 cm. Find the position and nature of the image.

Answer:

Given,

Distance of object (u) = – 10cm

Focal length = 15cm

Therefore, Distance of image = (v)

Using Mirror Formula for spherical mirror,

(1/ v)+ (1/u) = (1/f)

(1/v) + (1/-10cm) = (1/15cm)

=> (1/ v) = (1/15cm) + (1/10cm)

=> (1/v) = (2+3)/30) cm = (1/6) cm

=>v = 6cm

As distance of image is positive, this shows that image is formed at the other side of the mirror.

Now, magnification m = - (distance of image (v))/ (distance of object (u))

=> m = - (v/u)

=>m = (6cm)/ (-10cm) = +0.6cm.

As magnification is also positive this shows that image is erect.

Thus, Position of image = 6cm (behind the mirror, because distance of image is positive)

Nature of image = Virtual and erect (Because magnification is positive and image is formed behind the mirror)

**Question 13.**

The magnification produced by a plane mirror is +1. What does this mean?

Answer:

Magnification produced by a plane mirror is +1. This means that:-

- Size of the image is equal to the size of the object.
- (+) ive sign implies image is erect.

A plane mirror always forms virtual, erect and image of same size as of object.

**Question 14.**

An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm.

Find the position of the image, its nature and size.

Answer:

Given

Size of objet = 5.0 cm

Distance of object (u) = – 20cm

Radius of curvature (R) = 30cm

Thus, focal length = R/2 = 30cm/2 = 15cm

Distance of image = (v)

And magnification = (m)

Using, Mirror Formula, (1/v) + (1/u) = (1/f)

=> (1/v) = (1/-20cm) = (1/15cm)

=> (1/v) = (1/15cm) + (1/20cm)

=> (1/v) = (4+3)/ (60) cm = (7/60) cm

=>v= (60/7) cm= 8.57cm

Now, Magnification, m =-(v/u)

=> m =-(8.57cm)/ (-20cm) = (8.57/20)

=> m =0.43.

Now, m = (height of image) (h’)/ (height of object) (h)

=> 0.43 = (h’)/ (5.0)

=> h’=0.43 x 5.0cm

=> h’= 2.15cm.

Image is formed behind the mirror at a distance of 8.57cm,image formed is erect, virtual and of size 2.15cm.

** Question 15.**

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm.

At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained?

Find the size and the nature of the image.

Answer:

Given:-

Size of object (h) = 7.0 cm

Distance of object (u) = – 27cm

Focal length (f) = – 18cm (from concave mirror)

Distance of image = (v)

Height of image= (h')

Using, Mirror Formula,

(1/v)+ (1/u) = (1/f)

=> (1/v) + (1/-27cm) = (1/-18cm)

=> (1/v) =-(1/18) + (1/27)

=> (1/v) = (-3 +2)/ (54) cm= - (1/54) cm

=>v= -54cm

Now, magnification m = - (v/u)

=>m =-(-54cm)/ (-27cm) = - (54)/ (27) =-2

Also for size of image, m = (h’)/ (h)

Therefore, (-2) = (h’)/ (7cm)

=> Height of image h’ = -14cm.

Thus,

So screen should be placed at a distance of 54cm in front of mirror.

Image formed is twice the size of the object and it is real and inverted because both distance and magnification are negative.

**Question 16.**

Find the focal length of a lens of power – 2.0 D. What type of lens is this?

Answer:

Given,

Power of lens (P) = – 2.0 D

Thus, focal length = (f)

We know that, P = (1/f)

Or, – 2.0 D = (1/f)

Or, f = (1/– 2.0) D

Or, Focal length (f) = – 0.5 m

Since, power of length is (-ive), therefore it is a concave lens.

**Question 17.**

A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Answer:

Given:-

Power of lens (P) = + 1.5 D

Thus, focal length = (f)

We know that, P = (1/f)

Therefore (+ 1.5) D = (1/f)

Or, f = (1/1.5D) =0.66 m

Thus, focal length of the lens is equal to 0.66 m.

Power is (+) ive therefore the prescribed lens is a converging lens.

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