Self Studies

Ray Optics and Optical Test - 21

Result Self Studies

Ray Optics and Optical Test - 21
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    The real depth of a swimming pool that appears to be 10 m10\ m for a swimmer if the index of water is 1.331.33 is
    Solution

  • Question 2
    1 / -0
    In order to increase the angular magnification of a simple microscope, one should increase
    Solution
    When the image is formed at infinity
    m=Df=DPm=Df=DP
    when the image is formed at the near point
    m=(1+Df)=1+DP
  • Question 3
    1 / -0
    A beam of light is converging towards a point LL. A plane parallel plate of glass of thickness tt, refractive index μ\mu is introduced in the path of the beam. The convergent point is shifted by : 
    (assume near normal incidence).

    Solution
    When glass plate of thickness t is introduced, the optical path increases by tn\dfrac tn. 
    So the convergence point shifts by ttμt – \dfrac t\mu i.e. t[11μ]t\left[1 – \dfrac 1\mu\right] nearer. 
  • Question 4
    1 / -0
    The radius curvature for a convex lens is 40 cm40\ cm, for each surface. Its refractive index is 1.51.5. The local length will be
    Solution
    By formula 1f=(μ1)(1R11R2)\dfrac{1}{f}=(\mu-1) \left(\dfrac{1}{R_1}- \dfrac{1}{R_2} \right)
    =(1.51)(140+140)=0.5×120=140=(1.5-1) \left(\dfrac{1}{40}+ \dfrac{1}{40} \right) =0.5 \times \dfrac{1}{20} =\dfrac{1}{40}
    f=40 cm\therefore f=40\ cm
  • Question 5
    1 / -0
    The time taken by sunlight to cross a 5 mm5\ mm thick glass plate (μ=3/2)(\mu =3/2) is
    Solution
    As we know,
    t=μxc=32×5×1033×108=0.25×1010st=\dfrac {\mu x}{c}=\dfrac {\dfrac {3}{2}\times 5\times 10^{-3}}{3\times 10^8}=0.25\times 10^{-10}s
  • Question 6
    1 / -0
    Finger prints are observed by the use of 
    Solution
    Microscope is used for enlarging tiny objects or details which our eyes cannot detect. It is used to highly magnify the patterns and design of Finger prints and is widely used for Fingerprint tracing.
  • Question 7
    1 / -0
    A person using a lens as a simple microscope sees an 
    Solution
    A simple microscope is just a convex lens with object lying between optical centre and focus of the lens.
  • Question 8
    1 / -0
    A ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If μ\mu is the refractive index of the material of the prism, the maximum value of the base angle for which light is totally reflected from the hypotenuse is :
    Solution
    If  α\alpha  is maximum value of base angle for which light is totally reflected from the hypotenuse and  θ\theta  is the minimum value of angle of incidence at hypotenuse for total internal reflection.
    Then ,
    θ=sin1( 1μ) \theta =sin^{-1}\left (  \dfrac{1}{\mu}\right )

    α+θ=90o\alpha +\theta =90^o

    α=90osin1(1/μ)=cos1(1/μ)\Rightarrow \alpha = 90^o-sin^{-1}\left ( 1/\mu \right )=cos^{-1}\left ( 1/\mu \right )
  • Question 9
    1 / -0
    The refractive index of the material of a concave lens is μ\mu   and is placed in the medium of refractive index μ1\mu ^{1}. If μ1>μ\mu ^{1}>\mu , then which of the following ray diagram is correct:

    Solution
    Since, the tendency of a concave lens is to diverge a parallel beam of light falling on the lens surface, it would show exactly the opposite behaviour if placed in a medium of higher refractive index than the material of the lens.
    Thus, the correct option will be one in which a concave lens diverges a parallel beam of a light ray.
  • Question 10
    1 / -0
    A beam of parallel rays is incident on a transparent slab of refractive index 3\sqrt{3} making an angle 3030^{\circ} with the surface of the slab. If the width of incident beam of light is 1.732 mm, the width of refracted beam is:
    Solution
    Using μ1sini=μ2sin r\mu _{1}sini=\mu _{2}sin\ r
    1×sin60=3 sin r1\times sin60^{\circ}=\sqrt{3}\ sin\ r

    sin r=12sin\ r=\dfrac{1}{2}

    AB=1.732mmcosiAB=\dfrac{1.732mm}{cosi}

    Now,
    BB1=AB cos rBB^{1}=AB\ cos\ r

    =1.732cosi×cos r=\dfrac{1.732}{cosi}\times cos\ r

    =1.73212 ×32=3 mm=\dfrac{1.732}{\dfrac{1}{2}}\ \times \dfrac{\sqrt{3}}{2}=3\ mm

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now