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How will you use two identical glass prisms so that a narrow beam of white light incident on one prism emerges out of the second prism as white light ? Draw and label the ray diagram.
 

By placing second (identical) prism in an inverted position with respect to the first prism a narrow beam of white light incident on one prism emerges out of the second prism as white light

Ray diagram, 

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Safeer PP

The refractive index of a medium 'x' with respect to a medium 'y' is 2/3 and the refractive index of medium 'y'with respect to medium'z' is 4/3. Find the refractive index of medium 'z' with respect to medium 'x'. If the speed of light in medium 'x' is 3\times 10^{8}\; ms^{-1}, calculate the speed of light in medium'y'.
Option: 1
Option: 2
Option: 3
Option: 4

(I) Z/X=?

X/Y=2/3

Y=3X/2...(1)

Y/Z=4/3

Y=4Z/3...(2)

Equate RHS of (1) and (2)

3X/2=4Z/3

9X=8Z

Therefore, Z/X=9/8.

Hence, refrective index of medium Z w.r.t medium X is 9/8.

(II)Let speed of light in medium 'Y' be 'V'

Let speed of light in medium 'X' be 'C' i.e. 3*10^8m/s

So, Y/X=3/2...(3)

C/V=3*10^8/V...(4)

Equate RHS of (3) and (4)

3/2=3*10^8/V

So, 3V=3*10^8*2

Therefore, V=3*10^8*2/3

V=10^8*2

Therefore, V=2*10^8m/s

Hence, speed of light in medium Y is 2*10^8m/s

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Safeer PP

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Consider the following properties of virtual images: (A)    cannot be projected on the screen
(B)    are formed by both concave and convex lens
(C)    are always erect
(D)    are always inverted The correct properties are :
Option: 1 (A) and (D)
Option: 2 (A) and (B)
Option: 3 (A), (B) and (C)
Option: 4 (A), (B) and (D)

For virtual images, following options are correct- 

(A)    cannot be projected on the screen
(B)    are formed by both concave and convex lens
(C)    are always erect

Hence option C is correct 

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Safeer PP

A real image is formed by the light rays after reflection or reflection when they: (A)    Actually meet or intersect with each other.
(B)    Actually converge at a point.
(C)    appear to meet when they are produced in the backward direction.
(D)    appear to diverge from a point. Which of the above statement are correct?
Option: 1 (A) and (D)
Option: 2 (B) and (D)
Option: 3 (A) and (B)
Option: 4 (B) and (C)

A real image is formed by the light rays after reflection or reflection when they:

(A)    Actually meet or intersect with each other.
(B)    Actually converge at a point.

Hence statements A and B are correct. 

Option A is correct 

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Safeer PP

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When an object is kept within the focus of a concave mirror, an enlarged image is formed behind the mirror.The image is :
Option: 1 real
Option: 2 inverted
Option: 3 virtual and inverted
Option: 4 virtual and erect

The image is virtual and erect

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Safeer PP

The laws of reflection hold true for :  
Option: 1 plane mirrors only
Option: 2 concave mirrors only
Option: 3 convex mirrors only 
Option: 4 all reflecting surfaces

The laws of reflection hold true for all reflecting surfaces

Hence option D is correct 

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Safeer PP

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

Laws of refraction: 

  1. The incident ray, the normal to any refracting surface at the point of incidence, and the refracted ray all lie in the same plane called the plane of incidence or plane of refraction.
  2. The ratio of the sine of the angle of incidence to the angle of refraction is always constant. 

                           \frac{\sin i}{\sin r}=\mathrm{constant}

                          

    \frac{\sin i}{\sin r}=\frac{\mu _{2}}{\mu _{1}}=\frac{v_{1}}{v_{2}}=\frac{\lambda_{1}}{\lambda_{2}}

or
     $$ \mu _{1} \sin i=\mu _{2} \sin r $$

\frac{\sin(i)}{ \sin(r)}= \mu _{21}     

= refractive index of the second medium with respect to the first medium.

For a glass slab as shown in figure

i = angle of incidence

r1 = angle of reflection

\frac{sini}{sinr_1}=\frac{\mu_g}{\mu_{air}}

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Safeer PP

A student focused the image of a candle flame 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 lens = 50.0 cm

Position of the screen = 88.0 cm

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

(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 (iii) as said above.

Answer :

(i) Focal length = 38 ÷ 2 = 19 cm

(ii) object distance u = 50 – 31 = 19 cm

                In this case, object distance = focal length

                This means that images is formed at infinity.

(iii) The image formed will be virtual and erect.

(iv)

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infoexpert23

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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 lens and its power used by each of them?

Answer:

P = 1 / f, where f is in meter.

The unit of power is Diopter. Lens is convex in the first case and concave in the second case. Power of lens (first student) = +2 diopter

Power of lens (second student) = -2 diopter

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infoexpert23

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

 

Using,

 m = - v/u = h?/h?

⇒ - v/u = 1/3 h?/h?

⇒ u = - 3v

Using mirror formula,

1/v + 1/u = 1/f, we get

when f = + 20 cm

⇒ 1/v - 1/3v = 1/20

⇒ 3 - 1/3v = 1/20

⇒ 2/3v = 1/20

⇒ 3v = 2 × 20

⇒ v = 2 × 30/3

⇒ v = 40/3 .. (i)

u = - 3v

By putting v value, we get

⇒ u = - 3 × 40/3

⇒ u = - 40 cm.

Object Distance = - 40 cm

If we take f = - 20 cm, value of u will be + 40 cm which is not possible.

m = - v/v

⇒ m = - 40/3/- 40

⇒ m = + 1/3

⇒ m = 0.33 cm

This implies that the nature of the image is erect and diminished.

Nature of Image = Convex Mirror.

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