Analytical Questions: Light

 Numerical Questions: Light

Analytical questions typically involve critical thinking and problem-solving. These types of questions require applying theoretical concepts, using formulas, and applying problem-solving skills to analyze and deduce outcomes or behaviors.

Recall

Refractive Index:

The refractive index n of a medium is a dimensionless number that describes how light propagates through that medium. It is defined as the ratio of the speed of light in a vacuum c to the speed of light in the medium v:

n = c/n

n (Refractive Index): It indicates how much the light slows down when it enters the medium. The higher the refractive index, the slower the light travels through the medium.

c: The speed of light in a vacuum, approximately 3×10⁸ m/s

v: The speed of light in the medium.

Refractive indices can vary depending on the nature of the medium and the circumstances under which light is passing through it.

1. Absolute Refractive Index

The absolute refractive index of a medium is the ratio of the speed of light in a vacuum to the speed of light in that medium.

Formula: n = c/v

​where c is the speed of light in a vacuum, and v is the speed of light in the medium.

2. Relative Refractive Index

The relative refractive index is the ratio of the speed of light in one medium to the speed of light in another medium. It indicates how light will bend or change direction when transitioning between two different media.

Formula:  n₂₁ = v₁/v₂ = n₂/n₁

v₁, v₂ are the speeds of light in media 1 and 2, respectively, and n₁​and n₂​ are the absolute refractive indices of media 1 and 2.

For Spherical Mirrors

1.      Concave Mirror

         Focal Length (f) is Negative (as Focus is to the left of Pole)

     Converging Mirror

      Forms Real and Inverted Image. (Except when object is between F and P)

S. No.	Position of Object	Position of Image	Size of Image	Nature of Image
1	Object at Infinity	At the Focus	Highly Diminished	Real and Inverted
2	Object Beyond the Centre of Curvature	Between the Centre of Curvature and Focus	Diminished	Real and Inverted
3	Object at the Centre of Curvature or Focus	At the Centre of Curvature	Same Size	Real and Inverted
4	Object Between the Centre of Curvature and Focus	Behind the Centre of Curvature	Enlarged	Real and Inverted
5	Object at the Focus	At Infinity	Highly Enlarged	Real and Inverted
6	Object Between the Focus and the Pole	Behind the Mirror	Enlarged	Virtual and Erect

2.      Convex Mirror

       Focal Length (f) is Positive (as Focus is to the right of Pole)

       Diverging Mirror

        Always forms Virtual, Erect & Diminished Image.

Mirror Formula:

1/f = 1/u + 1/v

Magnification m = hi/ho = -v/u

m = 1 –> Object and Image equal in height

m > 1 -> Image is Enlarged

m < 1 -> Image is diminished

m is positive indicates Virtual and Erect Image

m is negative indicates Real and Inverted Image

Lens

1.      Convex Lens

Focal Length (f) is positive

Converging Lens

Forms Real and Inverted Image except when object is between F1 and O

Power is also positive

    

Position of the object	Position of the image	Relative size of the image	Nature of the image
At infinity	At focus F2	Highly diminished	Real and inverted
Beyond 2F1	Between F2 and 2F2	Diminished	Real and inverted
At 2F1	At 2F2	Same size	Real and inverted
Between F1 and 2F1	Beyond 2F2	Enlarged	Real and inverted
At focus F1	At infinity	Infinitely large or highly enlarged	Real and inverted
Between focus F1 and the optical center O	On the same side of the lens as the object	Enlarged	Virtual and erect

2.      Concave Lens

     Focal Length (f) is negative

     Diverging Lens

     Always forms Virtual, Erect and Diminished Image

     Power is also Negative

Lens Formula:

1/f = 1/v - 1/u

Magnification m = hi/ho = v/u

Solved Numerical:

1. The linear magnification produced by a spherical mirror is +1/3. Analyzing this value, state the 

    (i) type of mirror and 

    (ii) position of the object with respect to the pole of the mirror. 

    Draw any diagram to justify your answer.

Solution:

Given: m = + 1/3

Recall - m is positive indicates an Erect and Virtual Image and m < 1 (⸪ m = 1/3 ) indicate
              diminished image.

(i)  A magnification of +1/3 indicates an erect and diminished image. Only a Convex Mirror produces
      such an image. Because Concave Mirror also formed Erect and Virtual image when object is
      between Focus (F) and Pole (P), However it forms an enlarged image.

(ii) Position of object: Between infinity and the pole.

The diagram below shows that for any position of the object between ∞ and P, the convex mirror always forms an erect and diminished image.

2.  A spherical mirror produces a magnification of -1 on a screen placed at a distance of  50 cm  from
     the mirror.

    (i) Write the type of the mirror.
    (ii) Find the distance of the image from the object.
    (iii) What is the focal length of the mirror?

    (iv) Draw the ray diagram to show the image formation in this case.

Solution:

Given: m = - 1

Recall - m is negative indicates an Real and Inverted  Image and m = 1 (⸪ m = 1) indicates
              same size  image as that of the object. 

(i) The mirror is concave since it forms a real image. The case when an object is at center of
     curvature (C), then the same size image is formed at center of curvature (C).

(ii) A magnification of -1 implies that both the object and image are located at the center
      of curvature, C. The distance of the image from the object is zero.

(iii) The radius of curvature is 50 cm. So, the focal length is 25 cm (1/2 x radius of
        curvature).(⸪ R = 2f)

(iv) The image formation is shown below.

3. An object is placed at a distance of 25 cm away from a converging mirror of focal length 20 cm.

    Discus the effect on the nature and position of the image if the position of the object changes from 25
    cm to 15 cm. justify your answer without using mirror formula.

Solution:

If the object is placed at 25 cm in front of the concave mirror having focal length 20 cm, this means that the object is placed in between focus and the center of curvature of the mirror. If the object is placed in front of the center of curvature then the image will be formed beyond the center of curvature. So the image formed is a real image. The nature of the image will be  real & inverted and enlarged

If the object is placed at 15 cm in front of the concave mirror having focal length 20 cm, this means that the object is placed in between focus and the pole of the mirror i.e. 10 cm (25 - 15). An object placed between the pole and focus of a concave mirror forms a virtual & erect image. The nature of the image will be will be erect and enlarged.

1.     4. Write one similarity and one dissimilarity between image formed by plane mirror and convex mirror.

Similarity:- Both produce Virtual Image.

Dissimilarity:- Convex Mirror produces diminished image while plane mirror produce the Image of the same size as that of the object.

1.      5.     A student has three concave mirrors A, B and C of focal lengths 20 cm, 15 cm and 10 cm
       respectively. For each concave mirror he performs the experiment of image formation for three
       values of object distance of 30 cm, 10 cm and 20 cm.

Giving reason answer the following:

(a) For the three object distances, identify the mirror which will form an image equal in size to
      that of object. Find at least one value of object distance.

(b) Out of the three mirrors, identify the mirror which would be preferred to be used for shaving
      purpose.

(c) For the mirror B, draw ray diagram for image formation for any two given values of object
      distance.

Given:

f_A = 20 cm, f_B = 15 cm, f_C = 10 cm, u₁ = 30 cm, u₂ = 10 cm, u₃ = 20 cm

a. For same size image, object should be placed at Center of Curvature,  ⸪ R = 2 f, i.e. R₁, = 40
    cm, R₂ = 30 cm and R₃ = 20 cm from the given focal lengths.

So for Mirror A, Image will not of same size for these positions because no given object
distance is  40 cm.

For Mirror B, Image will be of same size for position u1 = 30cm ⸪ R₂ =30 cm

For mirror C, Image will be of same size for position u3 = 20cm ⸪ R₃ =30 cm

b. We need enlarged and erect image for shaving. For erect and enlarged image, Object
    should be placed between Pole and Focus. Face would be generally kept at more than
    10 cm distance from the pole, so mirrors A and B are suitable for shaving.

c. The required Ray diagrams are as follows:

6. It is desired to obtain an erect image of an object, using concave mirror of focal length of 12 cm.

    (i) What should be the range of distance of an object placed in front of the mirror?

    (ii) Will the image be smaller or larger than the object? Draw ray diagram to show the

         formation of image in this case,.

    (iii) Where will the image of this object be, if it is placed 6 cm in front of the mirror? Draw ray

          diagram for this situation to justify your answer. Show the position of pole, principal focus

          and the center of curvature in the ray diagram.

Solution:

(i) In concave mirror erect is formed only when the object is placed between pole and focus.
     So, object distance should be less than 12 cm.

(ii) The image would be larger than the object since an enlarge virtual and erect image is formed
       when object  is between Focus(F) and Pole (P) of a concave mirror.

(iii) Here u = -6 cm, f = -12 cm
       Using Mirror Formula
       1/f = 1/v + 1/u
       1 /−12 = 1 / v + 1 / −6
       v = 12 cm

     The image is real and inverted as it is formed in front of the mirror.

7.  A concave mirror produces an image of size n times that of the object and of focal length f. If the image
     is real then find the distance of the object from the mirror.

Solution:

Given that

Size of image = n × size of object

n = Size of image / size of object = magnification

Since the image is real, it must be inverted hence magnification will be negative,

m = -n

Let d is the distance of object then,

m = -v/u 

-n = -v / d

      or

      v = nd

      Therefore, the mirror formula:

      1 / f = 1/v + 1/u

      Becomes,

      1/f = 1/n x d + 1/d

      or

      1/f = 1/d (1/n + 1)

      or

      1/d = n/ f(n + 1)

     Therefore,

      d = f (n + 1)/ n

8. The object when placed at a distance of 60 cm from a convex mirror, then the magnification produced is

    1/2. Where the object should be placed to obtain a magnification of 1/3?

Solution:

      u = -60 cm

m = 1/2

So,

-v/u = 1/2

and

v/60 = 1/2

or

v = 30 cm

    Since, the mirror formula is:

1 / v + 1 / u = 1 / f

Therefore,

1 / 30 + 1 / (-60) = 1/f

1/f = (2-1) / 60 = 1 / 60

f = 60 cm

Now for magnification = 1 / 3,

– v / u = 1 / 3

or

v = – u / 3

 Using mirror formula

 1 / v + 1 / u = 1 / f

1 / (-u/3) + 1/ u = 1/ 60

-3/ u + 1/u = 1/60

-2/ u = 1/60

or

u = -120 cm

       The object should be placed at 120 cm in front of mirror to get magnification of 1/3.

9. 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. 

Solution:

Given:

ho = +5 cm, u = -25 cm and f = +10 cm (Lens converging)
using lens formula

1/f = 1/v – 1/u

1/+10 = 1/v – 1/-25

1/v = 1/10 - 1/25

     = 3/50

v = 50/3 = 16.67 cm

Also, Magnification m  = hi/ho = v/u

                                     = 50/3/-25

                                     = -2/3

The negative sign indicates that it is a Real and Inverted Image.

Now, m = hi/ho

         hi = m x ho

             = -2/3 x 5

             = -10/3 cm (The height is 10/3 cm or 3.33 cm in front of the mirror.

10.  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?

Solution:

        Given:  P = +1.5 D

         f = 1/P = 1/+1.5 = 0.67 m.

As the power of lens is (+ve), the lens is converging lens.

  11. AB and CD, two spherical mirrors, from parts of a hollow spherical ball with its center at O as shown in
        the diagram. If arc AB = 12 arc CD, what is the ratio of their focal lengths? State which of the two mirrors
        will always form virtual image of an object placed in front of it and why?

 

Solution:

Focal length of a mirror is given by

Focal length = Radius of curvature /2

Since both the mirrors have same radius of curvature, therefore focal length of the two mirrors will be same, i.e.

f1/f2 = 1/1

Since virtual image is always formed by convex mirror. The mirror AB will always form virtual image.

12. An object is placed at a distance of 2 times of focal length from the pole of the convex mirror,
      calculate the linear magnification.

Solution:

Let the Focal length of mirror = f

So, the object distance, u = -2f 

The formula to calculate image distance we use mirror formula as,

1 / v + 1 / u = 1 / f

Therefore,

1 / v + 1 / -2f = 1 / f 

             1 / v = 1 / f + 1 / 2f  

                      = 3 / 2f

or

v = 2f / 3

Magnification is given as, 

m = – v / u

= - (2f/3) / (-2f) 

= 1/3

13. The image of an object formed by a lens is of magnification -1. If the distance between the
       object and its image is 60 cm, what is the focal length of the lens? If the object is moved 20
       cm towards the lens, where would the image be formed? State reason and also draw a ray
       diagram in support of your answer.

Solution:

Image with magnification -1 means that - Negative sign indicates image is Real & Inverted, and m = 1 indicates the size of object is same as that of the Image.

Therefore, object is at 2F and the image is also at 2F on the other side of the lens.

Therefore, distance between the object and its image is –

    4f = 60 cm

=> f = 15 cm

Object distance 2f = 30 cm,

if the object is shifted towards the lens by 20 cm,

the new object distance = 30 cm - 20 cm = 10 cm.

As can be observed, The new object (10 cm) is distance is less than the focal length (15 cm), and the image formed in this case would be virtual, erect and will form on the same side as the object.

14. The refractive indices of glass and water with respect to air are 3/2 and 4/3 respectively. If speed of
       light in glass is 2 × 10⁸ m/s, find the speed of light in water.

Solution:

The refractive index n of a medium is given by:

 n = c / v

where:

c is the speed of light in a vacuum (approximately 3×10⁸ m/s),

      v is the speed of light in the medium.

    Given:

    Refractive index of glass, n_g  = 3/2​,

 Refractive index of water,    n_w = 4/3​,

         Speed of light in glass, v_g = 2×10⁸ m/s

We know the refractive index of glass is:

n_g = c/v_g ----- (1)

Now, to find the speed of light in water v_w, we use:

n_w = c/v_{w --------- (2)}

from (1) and (2), We have, ng/ n_w =  (c/v_g) / (c / v_w)

                                                 = (c x v_w) /(c x v_g)

                                                 = v_w / v_g

                       (3/2) /(4/3)         =  v_w/    (2 x 10⁸)

                                                  v_w   = (9 x 2 x 10⁸) /(8)

                                                         = 2.25 x 10⁸ m/s