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All questions of Curved Mirrors for Grade 9 Exam

A linear object is placed at a distance equal to focal length of a convex mirror. Its image is formed
  • a)
    at infinite distance
  • b)
    at the principal focus of mirror
  • c)
    behind the mirror at a distance f/2
  • d)
    in front of mirror at a distance f/2
Correct answer is option 'C'. Can you explain this answer?

When a linear object is placed at a distance equal to focal length of a convex mirror then its virtual image is form ed behind the mirror at a distance f/2 
As per sign convention u = - f and focal length o f convex mirror is + ve, hence from mirror formula we have 
  or   or 

Transparent medium is one :
  • a)
    Which allows light to pass through
  • b)
    Which absorbs most of the light
  • c)
    Which do not allows light to pass through
  • d)
    None of these
Correct answer is option 'A'. Can you explain this answer?

Imk Pathshala answered
A transparent medium is a material that partially allows light to pass through.
When light encounters a transparent medium, it can penetrate the material and transmit through it.
Examples of transparent mediums include glass, water, and air.
Transparent mediums are essential for various applications such as optics, windows, and lenses.

     
 

Can you explain the answer of this question below:

A concave lens of 20 cm focal length forms an image 15 cm from the lens. What is the object distance?

  • A:

    60 cm

  • B:

    30 cm

  • C:

    -60 cm

  • D:

    -30 cm

The answer is c.

Pooja Shah answered
Focal length= -20(as it is concave lens)
v= -15 (as concave lens always forms virtual and erect image on left of lens)
Putting these values in lens formula,
1/ -20 - 1/u = 1/ -15
-1/ u= 1/-15 + 1/20
-1/u = -4+3/60
-1/u = -1/60
-u = -60
[u =60]

An object 2 cm high is placed at a distance of 15 cm from a concave mirror which produces an inverted image 4 cm high. Find the position of the image.
  • a)
    -30 cm
  • b)
    30 cm
  • c)
    40 cm
  • d)
    -40 cm
Correct answer is option 'B'. Can you explain this answer?

Krishna Iyer answered
The magnification produced by a concave mirror is given by the formula:
m = -v/u
where, m = magnification v = image distance u = object distance
From the problem, we have: u = -15 cm (since the object is placed in front of the mirror, the distance is negative) m = -4/2 = -2 (since the image is inverted and 4 cm high whereas the object is 2 cm high)
Substituting these values in the magnification formula, we get:
-2 = -v/(-15)
Simplifying, we get:
v = 30 cm
Therefore, the position of the image is 30 cm from the concave mirror.

The lens which is used to correct myopia (shortsightedness) is
  • a)
    Both convex and concave
  • b)
    Concave lens
  • c)
    Converging lens
  • d)
    Convex Lens
Correct answer is option 'B'. Can you explain this answer?

Raghav Bansal answered
Shortsightedness is corrected using a concave (curved inwards) lens which is placed in front of a myopic eye, moving the image back to the retina and making it clearer.

A virtual, erect and magnified image of an object is to be produced with a concave mirror of focal length 12 cm. Object may be placed at a distance of
  • a)
    10 cm from the mirror
  • b)
    15 cm from the mirror
  • c)
    24 cm from the mirror
  • d)
    48 cm from the mirror
Correct answer is option 'A'. Can you explain this answer?

Arun Sharma answered
A concave mirror forms a virtual, erect and magnified image when an object is placed between pole and focus point of the mirror. As focal length of given concave mirror, hence object must be placed at a distance less than 12 cm i.e., u < 12 cm. Thus, the object may be placed at a distance of 10 cm from the mirror.

A ray of light AM is incident on a concave mirror as shown below. Then which of the following ray diagrams is correct for the reflected ray ?
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Naman pandey answered
As the incident light ray is coming parallel to the principal axis of given concave mirror, the reflected ray must pass through the principal focus of mirror.

A lens has a power of +0.5 D. It is
 
  • a)
    a concave lens of focal length 5 m
  • b)
    a convex lens of focal length 5 cm
  • c)
    a convex lens of focal length 2 m
  • d)
    a concave lens of focal length 2 m
Correct answer is option 'C'. Can you explain this answer?

Sanya jain answered
Given, power of the lens = 0.5 D.

We know that the power of the lens is given by the formula:

Power (P) = 1/f, where f is the focal length of the lens.

Therefore, we can write:

0.5 D = 1/f

Solving for f, we get:

f = 1/0.5 D

f = 2 m

Hence, the focal length of the lens is 2 m.

Therefore, the correct option is (c) a convex lens of focal length 2 m.

Explanation:

A convex lens has a positive power and can converge the light rays to a point. The focal length of a convex lens is positive. When the power of the lens is given, we can find the focal length of the lens using the formula P = 1/f. Here, the power is given as 0.5 D. On substituting the values, we get the focal length as 2 m. Hence, the lens is a convex lens of focal length 2 m.

Which lens always forms diminished and erect image ?
  • a)
    Convex lens
  • b)
    Concave lens
  • c)
    Converging lens
  • d)
    Both convex and concave
Correct answer is option 'B'. Can you explain this answer?

A concave lens (also known as a diverging lens) is thinner in the center and thicker at the edges. When light rays pass through a concave lens, they diverge (spread out), causing the rays to appear to come from a single point on the same side of the lens as the object. This results in the formation of a virtual image.
Key characteristics of the image formed by a concave lens:
  • Diminished: The image is smaller than the actual object.
  • Erect: The image is upright, meaning it has the same orientation as the object.
  • Virtual: The image cannot be projected on a screen because the light rays do not actually meet but only appear to do so when extended backward.
Because of these properties, a concave lens always forms a diminished, erect, and virtual image, no matter where the object is placed in front of the lens.
On the other hand, a convex lens (also known as a converging lens) can form different types of images (real and inverted or virtual and erect) depending on the position of the object relative to the lens. But it does not always form a diminished and erect image, unlike the concave lens.

The unit of linear magnification is
  • a)
    Dioptre
  • b)
    Metre
  • c)
    Metre per second
  • d)
    It is unitless
Correct answer is option 'D'. Can you explain this answer?

Gaurav Kumar answered
Magnification is thr ratio of size of image to size of object and since it is a ratio of tel similar quantities hence it is unitless. 

In a convex lens, where is the image formed, when an object is placed at 2F ?
  • a)
    Between F and 2F
  • b)
    At focus (F)
  • c)
    At 2F on the other side
  • d)
    At 2F on the same side
Correct answer is option 'C'. Can you explain this answer?

Gaurav Kumar answered
At the 2F point, the object distance equals the image distance and the object height equals the image height. As the object distance approaches one focal length, the image distance and image height approaches infinity.

If the magnification has a negative sign, the image formed by the concave mirror must be
  • a)
    Real and inverted
  • b)
    Virtual and inverted
  • c)
    Virtual and erect
  • d)
    Real and erect
Correct answer is option 'A'. Can you explain this answer?

Naina Sharma answered
We know that, if the magnification value is negative sign in the concave mirror, then the image will be real and inverted. Especially, when you come to concave mirror, the images are formed at the left of the mirror. So, it forms real and inverted.

No matter how far or close you stand from a mirror, your image is always virtual and erect. The mirror is
  • a)
    convex mirror
  • b)
    plane mirror
  • c)
    concave mirror
  • d)
    either a convex or a plane mirror
Correct answer is option 'D'. Can you explain this answer?

Raghav Bansal answered
Irrespective of the position of an object both convex mirror and plane mirror form virtual and erect images. Of course image formed by a plane mirror is of same size as the object but image formed by convex mirror is always diminished one.

The muscles of the iris control the
  • a)
    focal length of the eye-lens
  • b)
    opening of the pupil
  • c)
    shape of the crystalline lens
  • d)
    optic nerve
Correct answer is option 'B'. Can you explain this answer?

Nk Classes answered
The Laws of Reflection for Mirrors

- The laws of reflection apply to all mirrors, regardless of their shape.
- These laws state that the angle of incidence is equal to the angle of reflection.
- This means that the light rays that strike the mirror are reflected at the same angle.
- Whether the mirror is concave, convex, or plane, these laws will always hold true.
- Therefore, option B is correct as it includes all mirrors, not just a specific type.

The angle of incidence of any light ray passing through the centre of curvature of a spherical mirror is
  • a)
  • b)
    45°
  • c)
    90°
  • d)
    60°
Correct answer is option 'A'. Can you explain this answer?

Chirag raman answered
The angle of incidence of any light ray passing through the centre of curvature of a spherical mirror is 0°, because a line joining centre of curvature to any point on the mirror is a normal drawn at that point of the mirror.

Rays from sun converge at a point 15 cm in front of a concave mirror. Where should an object be placed so that size of the image is exactly equal to the size of the object ?
  • a)
    30 cm in front of mirror
  • b)
    15 cm in front of mirror
  • c)
    Between 15 cm and 30 cm in front of mirror
  • d)
    Less than 15 cm in front of mirror
Correct answer is option 'A'. Can you explain this answer?

Kiran Mehta answered
As light rays from sun (u = ) converge at a point 15 cm in front of a concave mirror, hence focal length of concave mirror f = - 15 cm.
To form an image of exactly same size as that of an object, the object should be placed at the centre of curvature (u = R = 2f) of mirror. Hence, the object should be placed at 30 cm in front of mirror.

The angle of incidence is the angle between
  • a)
    the incident ray and the surface of the mirror
  • b)
    the reflected ray and the surface of the mirror
  • c)
    the normal to the surface and the incident ray
  • d)
    the normal to the surface and the reflected ray
Correct answer is option 'C'. Can you explain this answer?

The angle between the incident ray and the normal is called as angle of incidence and it is generally denoted by i. The angle between the normal and the reflected ray is called as angle of reflection and it is generally denoted by r.

In which of the following, the image of an object placed at infinity will be highly diminished and point sized ?
  • a)
    Concave mirror only
  • b)
    Convex mirror only
  • c)
    Convex lens only
  • d)
    All types of mirrors and lenses
Correct answer is option 'D'. Can you explain this answer?

Flembe Academy answered
  • When an object is placed at infinity, a parallel beam of light is incident on the mirror/lens.
  • For all types of mirrors and lenses the beam will get focussed at the principal focus of mirror/lens and a highly diminished, point size image is formed there.

The minimum distance between an object and its real image in a convex lens is (f = focal length of the lens) 
  • a)
    2.5 f
  • b)
    2 f
  • c)
    4 f
  • d)
    f
Correct answer is option 'C'. Can you explain this answer?

Arun Sharma answered
Let the distance between the object and its real image formed by convex lens be d1​.
Let the distance of the object from lens be x,so,the image distance from the lens is (d-x)
The minimum distance between an object and its real image in a convex lens is 4f.
hence,option C is correct.
.

Refractive index of glass w.r.t. air is 3/2. What is the refractie index of air w.r.t glass ?
  • a)
    2/3
  • b)
    1
  • c)
    Zero
  • d)
    (3/2)2
Correct answer is option 'A'. Can you explain this answer?

Zara Khan answered

If  refractive index of glass w.r.t air is 3/2 , then  refractive index of air w. r. t glass will be it's reciprocal ie. ⅔ ..

An object is placed before a convex lens. The image formed
  • a)
    is always real
  • b)
    may be real or virtual
  • c)
    is always virtual
  • d)
    is always erect
Correct answer is option 'B'. Can you explain this answer?

Convex Lens: A convex lens is thicker at the center and thinner at the edges. When light rays pass through a convex lens, they converge at a point called the focal point.
Image Formation: When an object is placed before a convex lens, the light rays from the object refract through the lens and form an image on the other side. The image formation by a convex lens depends on the distance of the object from the lens.
Real or Virtual Image: The image formed by a convex lens can be real or virtual, depending on the position of the object relative to the lens.
Real Image: A real image is formed when the light rays actually converge at a point after passing through the lens. This real image can be projected onto a screen and is always inverted.
  • Virtual Image: A virtual image is formed when the light rays appear to converge at a point on the same side of the lens as the object. This virtual image cannot be projected onto a screen and is always upright.
  • Conclusion: Therefore, when an object is placed before a convex lens, the image formed may be real or virtual, depending on the position of the object relative to the lens.

     
 

An object 4 cm tall is placed in front of a convex lens. It produces an image 3 cm tall. What is the magnification of the lens ?
  • a)
    1.33
  • b)
    12
  • c)
    0.75
  • d)
    11
Correct answer is option 'C'. Can you explain this answer?

Rohit Sharma answered
We know, height of the object is 4 cm h1, height of the image is 3 cm, h2. 
So we have, m = h2/h1 
=> m = 3/4  
=> m = 0.75
Therefore, magnification of the lens is 0.75 

A point object is placed on the principal axis of a spherical mirror. The object-distance u is
 
  • a)
    definitely negative
  • b)
    definitely positive
  • c)
    positive if the object is to the left of the centre of curvature
  • d)
    positive if the object is to the right of the centre of curvature
Correct answer is option 'A'. Can you explain this answer?

Ritu Saxena answered
Option ( a) is the correct answer. As the object is always placed on the left side of the mirror and according to the sign convention, it has negative value for 'so axis.
Therefore, spherical mirrors have only one reflecting surface and it will be negative only.

An object is placed at the centre of curvature of a concave mirror. The distance between its image
and the pole is
 
  • a)
    equal to f
  • b)
    between f and 2f
  • c)
    equal to 2f
  • d)
    greater than 2f
Correct answer is option 'C'. Can you explain this answer?

Ishaani mehta answered
Explanation:

The given situation can be represented as follows:

![image.png](attachment:image.png)

Here, C is the centre of curvature of the concave mirror, F is the focus, and P is the pole.

Now, let us consider the path of light rays from the object to the mirror and then to its image.

- The light rays from the object are parallel to the principal axis and fall on the mirror.
- At the point of incidence, the light rays are reflected and converge towards the focus F.
- However, since the object is placed at the centre of curvature C, the reflected rays pass through F and become parallel to the principal axis.
- These parallel rays then converge at the position of the image I.

Therefore, we can see that the image is formed at a distance equal to twice the focal length from the mirror.

- From the mirror formula:

1/f = 1/v + 1/u

where f is the focal length, v is the image distance, and u is the object distance.

- In this case, since the object is placed at C, u = -2f (negative sign indicates that it is on the opposite side of the mirror).
- We want to find v, the distance between the image and the pole.
- Substituting the given values, we get:

1/f = 1/v - 1/2f

Multiplying both sides by vf2, we get:

2f = v + f

Therefore, v = 2f - f = f.

Hence, the distance between the image and the pole is equal to the focal length of the mirror, which is option C.

In which of the following, the image of an object placed at infinity will be highly diminished and point sized ?
  • a)
    Concave mirror only
  • b)
    Convex mirror only
  • c)
    Convex lens only
  • d)
    Concave mirror, convex mirror, concave lens and convex lens
Correct answer is option 'D'. Can you explain this answer?

Krishna Iyer answered
The incident ray coming from the object placed at infinity will be parallel to the principal axis. When the parallel beam of light incident on a mirror or lens, irrespective of their nature, after reflection/refraction, will pass or appear to pass through their principal focus. Hence highly diminished and point size image will be formed at their focus.

In a convex mirror for any position of the object in front of mirror, the image formed is
  • a)
    situated between pole and principal focus of mirror
  • b)
    virtual and erect
  • c)
    diminished in size
  • d)
    all of these
Correct answer is option 'D'. Can you explain this answer?

For a convex mirror irrespective of the position of object the image formed is virtual, erect and diminished one. Moreover, the image is always situated between pole and principal focus of the mirror.

The distance at which an object should be placed from a thin convex lens of focal length 10 cm to obtain a virtual image of double of its size is​
  • a)
    15 cm
  • b)
    – 5 cm
  • c)
    -10cm
  • d)
    5.5 cm
Correct answer is option 'B'. Can you explain this answer?

Neha Patel answered
given : f = 10 cm , m = 2

v / u = 2

v = 2u 

v = 2u -------(1)

according to lens formula , 

1/v - 1/u = 1/f

1/v - 1/u = 1/10 --------(2)

substitute (1) in (2)

1/2u - 1/u = 1/10

1 - 2 / 2u = 1/10

-1 / 2u = 1/10

-10 = 2u

u = -10/2

u = -5 cm

therfore object should be placed 5 cm away from the lens

A convergent beam of light passes through a diverging lens of focal length 0.2 m and comes to focus 0.3 m behind the lens. Find the position of the point at which the beam would converge in the absence of lens.​
  • a)
    0.1 m
  • b)
    0.22 m
  • c)
    0.2 m
  • d)
    0.12 m
Correct answer is option 'D'. Can you explain this answer?

Neha Patel answered
The image is 0.3 m to the right of the lens. This is 0.5 m TO THE RIGHT OF the left hand (image) focal point. 
As explained above, concave lenses have their focal points on the opposite side of the lens to a convex lens. 

So in the formula so* si = f^2 si = 0.5 
f^2 = 0.2 ^2= 0.04 

Therefore so = 0.04/ 0.5 = 0.08 m 

This means that the object must have been 0.08 m to the left of its focal point. 

which places it at 0.12m from the lens 

This would be the point at which the rays of light would have converged if the lens had not been present. 

It is actually a VIRTUAL OBJECT because the light never actually reaches this point. 
The final point the light reaches is a REAL IMAGE because the light actually does focus at this point as can be seen by putting a paper screen there.

Drop of water behaves likes a
  • a)
    Diverging lens
  • b)
    Concave lens
  • c)
    Convex lens
  • d)
    Both convex and concave
Correct answer is option 'C'. Can you explain this answer?

Krishna Iyer answered
The surface of a water drop curves outward to make a dome. This outward, or convex, curvature bends light rays inward.  will act as a concave lens that bends the light rays outward. As a result, letters seen through the layer of water in a cup appear smaller than they are.

A magnifying glass is a
  • a)
    Both convex and concave
  • b)
    Concave lens
  • c)
    Diverging lens
  • d)
    Convex Lens
Correct answer is option 'D'. Can you explain this answer?

A magnifying glass is a convex lens used to make an object appear much larger than it actually is. This works when the object is placed at a distance less than the focal length from the lens.

Which lens always forms a virtual image ?
  • a)
    Concave lens
  • b)
    Convex lens
  • c)
    Converging lens
  • d)
    Both convex and concave
Correct answer is option 'A'. Can you explain this answer?

Nandini nair answered
Explanation:

A virtual image is an image that cannot be projected onto a screen. It is always located on the opposite side of the lens from the object and is always upright. Virtual images are formed when light rays appear to come from a point behind the lens, rather than actually converging at that point.

Concave Lens:
- A concave lens is thinner at the center and thicker at the edges.
- When light rays pass through a concave lens, they diverge or spread out.
- As a result, a concave lens always forms a virtual image that is upright, smaller in size, and located between the lens and the object.

Convex Lens:
- A convex lens is thicker at the center and thinner at the edges.
- When light rays pass through a convex lens, they converge or come together.
- Depending on the placement of the object, a convex lens can form either a real or virtual image.

Converging Lens:
- A converging lens is another name for a convex lens.
- As discussed above, a converging lens can form both real and virtual images.

Conclusion:
Based on the above explanation, it can be concluded that a concave lens always forms a virtual image.

Which of the following can be used to form a virtual image of an object?
I. convex lens
II. concave lens
III. concave mirror
  • a)
    II only
  • b)
    II and III only 
  • c)
    I and III only
  • d)
    I, II and III
Correct answer is option 'D'. Can you explain this answer?

Adidev Nair answered
A concave lens forms
 real and inverted image
. When the 
object
 is placed very close to the 
lens
, the 
image
 formed is 
virtual
, erect and magnified. 

 A 
concave lens
 always 
forms
 erect, 
virtual
 and smaller 
image
 than the 
object
.

Concave
 mirrors 
can produce
 both real and 
virtual images
; they 
can
 be upright (if 
virtual
) or inverted (if real); they 
can
 be behind the 
mirror
 (if 
virtual
) or in front of the 
mirror
 (if real); they 
can
 also be enlarged, reduced, or the same size as 
object
.


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