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Can someone help me with this, Pick the wrong answer in the context with rainbow

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 

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S solutionqc
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 
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Medical
I need help with - Optics - NEET-2

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 

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P Plabita
Option 1)2 : 1 Option 2)1 : 2 Option 3)2 : 3 Option 4)3 : 4 
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Medical
Tell me? 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 ?

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)

    0 

  • Option 3)

    equal to angle of incidence 

  • Option 4)

    90 

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P Plabita
For internal refection  at Option 1)180 Option 2)0 Option 3)equal to angle of incidence Option 4)90 
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Medical
Example 9.9 from ncert....In the refraction from 2nd lens we use 1/v1-1/10 =1/(-10) but now the image is acting like a virtual object..t

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

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A Abhishek Sahu
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
Medical
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Need explanation for: - Optics - NEET

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 5000{\text \AA}  is incident on the slit. The first secondary maximum is observed at an angle of:

  • Option 1)

    {\text{sin}}^{-1}\left(\frac{1}{4}\right)

  • Option 2)

    {\text{sin}}^{-1}\left(\frac{2}{3}\right)

  • Option 3)

    {\text{sin}}^{-1}\left(\frac{1}{2}\right)

  • Option 4)

    {\text{sin}}^{-1}\left(\frac{3}{4}\right)

 

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P prateek
As we learnt in Fraunhofer Diffraction -   - wherein A condition of nth minima. slit width angle of deviation   For first minima;  since  For  secondary maxima Option 1) Incorrect Option 2) Incorrect Option 3) Incorrect Option 4) Correct
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Medical
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Help me please, - Optics - NEET

The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio

\frac{{I_{\max } - I_{\min } }} {{I_{\max } + I_{\min } }}

will be

  • Option 1)

    \frac{{\sqrt n }} {{n + 1}}

  • Option 2)

    \frac{{2\sqrt n }} {{n + 1}}

  • Option 3)

    \frac{{\sqrt n }} {{\left( {n + 1} \right)^2 }}

  • Option 4)

    \frac{{2\sqrt n }} {{\left( {n + 1} \right)^2 }}

 

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S shivalik
As we learnt in Minimum amplitude & Intensity -   - wherein     Maximum amplitude & Intensity - When   - wherein           Option 1) This is incorrect option Option 2) This is correct option Option 3) This is incorrect option Option 4) This is incorrect option
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Medical
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Need clarity, kindly explain! The angle of incidence for a ray light at a refracting surface of a prism is45°. The angle of prism is 60°. If the ray suffers minimum deviation throughthe prism, the angle of minimum deviation and refractive ind

The angle of incidence for a ray light at a refracting surface of a prism is 45°. The angle of prism is 60°. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are:

  • Option 1)

    45°; \frac{1}{\sqrt{2}}

  • Option 2)

    30°;\sqrt{2}

  • Option 3)

    45°;\sqrt{2}

  • Option 4)

    30°;\frac{1}{\sqrt{2}}

 

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P Plabita
As we learnt in Condition of maximum deviation -     - wherein       i =45 A=60 for minimum deviation = = Option 1) 45°;  Incorrect option Option 2) 30°; Correct option Option 3) 45°; Incorrect option Option 4) 30°; Incorrect option
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Medical
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Help me answer: 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

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

  • Option 1)

    8

  • Option 2)

    10

  • Option 3)

    12

  • Option 4)

    16

 

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P prateek
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=1.5x5=7.5 cm  apparent depth=3cm from other side   Real depth from other side =3x1.5cm=4.5  Thickness of...
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Medical
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Help me solve this A linear aperture whose width is 0.02 cm is placed immediately in front of alens of focal length 60 cm. The aperture is illuminated normally by aparallel beam of wavelength 5 x 10-5 cm. The distance of the first darkband of the dif

A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a
parallel beam of wavelength 5 x 10-5 cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is

  • Option 1)

    0.10 cm

  • Option 2)

    0.25 cm

  • Option 3)

    0.20 cm

  • Option 4)

    0.15 cm

 

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P Plabita
None The screen must be placed at a distance of 60cm from lens. Hence  For first d ark b and b = 1.5 mm or 0.15cm Option 1) 0.10 cm This solution is incorrect. Option 2) 0.25 cm This solution is incorrect. Option 3) 0.20 cm This solution is incorrect. Option 4) 0.15 cm This solution is correct.
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