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Pick the wrong answer in the context with rainbow

• Option 1)

When the light rays undergo two internal  reflection in a water drop , a secondary rainbow is formed

• Option 2)

The order of colors is reversed in the secondary rainbow

• Option 3)

An observer can see a rainbow when his front is towards the sun

• Option 4)

Rainbow is a combined effect of dispersion , refreaction and reflection of sunlight

Rainbow can seen when back is towards the Sun.Option 1)When the light rays undergo two internal  reflection in a water drop , a secondary rainbow is formed Option 2)The order of colors is reversed in the secondary rainbow Option 3)An observer can see a rainbow when his front is towards the sun Option 4)Rainbow is a combined effect of dispersion , refreaction and reflection of sunlight

Two similar thin equiconvex lenses , of focal length f each , are kept  coaxially in contact  with each other such that the focal length of the combination is $F_1$ . When the space between the two lenses is filled  with glycerin  then the equivalent  focal length is $F_2$ . The ratio $F_1 : F_2$

• Option 1)

2 : 1

• Option 2)

1 : 2

• Option 3)

2 : 3

• Option 4)

3 : 4

Option 1)2 : 1 Option 2)1 : 2 Option 3)2 : 3 Option 4)3 : 4

In total internal reflection when the angle of incidende is equal to the critical angle for the pair of media in contact  what will be angle of refraction ?

• Option 1)

180

• Option 2)

• Option 3)

equal to angle of incidence

• Option 4)

90

For internal refection  at Option 1)180 Option 2)0 Option 3)equal to angle of incidence Option 4)90

Convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d is:-

Refraction through parallel slab -   - wherein   shifting of object from slab  thickness of slab  Refractive Index of slab. F =  = 5 cm Due to glass plate shift s = t  = 1.5  = 0.5 New u =9.5 cm Using lens formula =  v = 10.55 cm Shift  is 0.55 cm away from the lens.

A man of height 'h' is walking away from a street lamp with constant speed 'v'.The height of the street lamp is 3h.The rate at which of the length of the man's shadow is increasing when he is at a distance of 10h from the base of the street lamp is

a)v/2

b)v/3

c)2v

d)v/6

form two triangles shadow triangle and men . these two will be similar  Now try to solve.

The reflecting surface of a plane mirror is vertical. A particle is projected in a vertical plane which is also perpendicular to plane of the mirror . The initial velocity of the particle is 10m/s and the angle of projection is 60 degree.The point of projection is at a distance 5m from the mirror .The particle moves towards the mirror .Just before the particle touches the mirror the velocity of approach of the particle and its image is

a)10m/s

b)5m/s

c)10$\sqrt{}$3 m/s

d)5$\sqrt{}$3 m/s

any instant only the component of velocity perpendicular to the mirror matters as parrallel component will be same for both image and object . the perpendicular component is constant= ucos60=5m/s

so the object and image are moving towards each other at a speed of 5m/s

the velocity of approach=10m/s

2)A body of mass accelerates uniformly from rest   to $v_{1}$  in time $t_{1}$  The instantaneous power delivered to the body as a function of time t is

answer for this question is given below

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There are two plane mirror with reflecting surfaces facing each other .Both the mirrors are moving away from each other .A point object is placed between the mirrors. The velocity of the image formed due to nth reflection will be

a)nv

b)2nv

c)3nv

d)4nv

There are two plane mirror with reflecting surfaces facing each other .Both the mirrors are moving away from each other .A point object is placed between the mirrors. The velocity of the image formed due to nth reflection will be As you can see the first image formed is at a distance equal to 2 times of the distance between the mirrors. Similarly the distance between nth order image will be 2n...
Q10.jpg By what angle shoulcl M. be rotated, so that the light ray after reflection froin both the mirror become horizontal? (as shown in figure)
To become horizontal angle of reflection from first mirror is 50 degree now the ray should make same angle with M2 if its horizontal. But M2 is tilted by 25 degree. So with respect to M2 normal i=50+25=75 degree. Now if the ray is horizontal then the angle between the emerging ray and mirror M2 is 90-25=65 so mirror has to be rotated (75-65)/2 = 5 degree counterclock wise to make emerging ray...

Two plane mirrors are inclined at 70 degree. A ray incident on one mirror at angle q after reflection falls on the second mirror and is reflected from there parallel to the first mirror ,q is

a)50 degree

b)45 degree

c)30 degree

d)55 degree

Angle q can be calculated by using the below formula:
15508342796514206909676845011232.jpg 15508343111324497254251767481204.jpg

An air bubble in a glass slab with refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is

@Krupa As we learnt in Relation between real depth and apparent depth -   - wherein = Refractive index of medium of incident. = Refractive index of medium of refraction. d = distance of object. = apparent  depth     Apparent depth = 5cm  from one side   Real depth=1.5  5=7.5 cm  apparent depth = 3cm from other side   Real depth from other side =3  1.5cm = 4.5 cm  Thickness of slab = 7.5cm...
15470918876053433028456896495246.jpg An unnumbered wall clock shows time 04:25:37 where 1st represents hours 2nd represents minutes and the last represents seconds. What time will its image in a plane mirror show
@Preethi 11 : 60 : 60 - 04 : 25 : 37 ___________ 07 : 35 : 25 the time is  07 : 35 : 25
Screenshot_730.png

For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index:

Here In the given situation A can varies from 0 to 90. lies between and 2.
In a diffraction pattern due to a single slit of width 'a', the first minimum is observed at an angle 30° when light of wavelength   is incident on the slit. The first secondary maximum is observed at an angle of:

For first minima;  since  For  secondary maxima

...so why is u=+10 and not -10 ? As it will be in the direction opposite to the incident rays

According to the setup the image formed by 1st lens is real and it is 10 cm from the 2nd lens. But the rays after having refraction from 1st lens form a virtual image and hence this virtual image is on the left side of the 2nd lens and that is why they took -10 and not +10 for u2
IMG-20181213-WA0007.jpg

A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance x from the source. The beam is reflected back as a spot on a scale placed just above the source L. When the mirror is rotated through a small angle , the spot of the light is found to move through a distance y on the scale. The angle is given by :

When mirror is rotated by an angle  then reflected ray will be rotated by an angle
Medical
123 Views   |

An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be

• Option 1)

30 cm towards the mirror

• Option 2)

36 cm away from the mirror

• Option 3)

30 cm away from the mirror

• Option 4)

36 cm towards the mirror

As we have learned Mirror Formula -   - wherein Object distance from pole of mirror. Image distance from pole of mirror. focal length of the mirror.   2nd case  Image gets displaced by 36 cm , away from the mirror           Option 1) 30 cm towards the mirror This is incorrect Option 2) 36 cm away from the mirror This is correct Option 3) 30 cm away from the mirror This is...
Medical
107 Views   |

The refractive index of the material of a prism is $\sqrt{2}$  and the angle of the prism is 30o. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is

• Option 1)

30o

• Option 2)

45o

• Option 3)

60o

• Option 4)

zero

As we have learned Angle of deviation through prism - - wherein angle of incidence angle of emergence angle of prism             Option 1) 30o This is incorrect Option 2) 45o This is correct Option 3) 60o This is incorrect Option 4) zero This is incorrect
Medical
205 Views   |

An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of

• Option 1)

large focal length and large diameter

• Option 2)

large focal length and small diameter

• Option 3)

small focal length and large diameter

• Option 4)

small focal length and small diameter

As we have learned Astronomical Telescope -   - wherein = focal length of objective focal length of eyepiece   for large angular magnification and high angular resolution , the objective lens of refracting telescope should have large diameter and large focal length         and resolving power        Option 1) large focal length and large diameter This is correct Option 2) large focal...
Medical
101 Views   |

In Young's double slit experiment the separation
d between the slits is 2 mm, the wavelength $\lambda$ of the light used is 5896 $A^0$ and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same   $\lambda$ and D) the separation between the slits needs to be changed to

• Option 1)

2.1 mm

• Option 2)

1.9 mm

• Option 3)

1.8 mm

• Option 4)

1.7 mm

As we have learned Young Double Slit Experiment - - wherein Distance of a point on screen from central maxima Path difference at that point    For nth maxima                  Option 1) 2.1 mm Option 2) 1.9 mm Option 3) 1.8 mm Option 4) 1.7 mm
Medical
80 Views   |

Unpolarised light is is incident from air on a plane surface of a material of refractive index $\mu$. At a particular angle of incidence 'i', it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation ?

• Option 1)

$i = sin^{-1}\left (\frac{1}{\mu} \right )$

• Option 2)

Reflected light is polarised with its electric vector perpendicular to the plane of incidence

• Option 3)

Reflected light is polarised with its electric vector parallel to the plane of incidence

• Option 4)

$i = tan^{-1}\left (\frac{1}{\mu} \right )$

As we have learned Relation between angle of incidence and angle of refaction -   - wherein refractive index of medium of incidence.   refractive index of medium where rays is refracted. angle of incidence. angle of refraction.    Apply snells law  also  or  also reflected light is polarized with its electric field perpendicular to plane of incidence         Option 1) This is...
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