Practice free Optics (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
A silicon optical fibre with a care diameter large enough has a care refractive inder of 1.50 and cladding refractive inder 1.47. The critical angle at the care cladding interface is
What is the working principal of optical fibre ?
One microammeter has a resistance of $100 \Omega$ and a full scale range of $ 50 \mu A$ . It can be used as a voltmeter or as a higher range ammeter provided a resistance is added to it. Pick the correct range and resistance combinations (s).
$ I_g = { V \over (R+G) } $
A convex lens of focal length f is placed somewhere in between an object and a screen. The distance between the object and the screen is x. If the numerical value of the magnification product by the lens is m, What is the focal length of the lens ?
x = u + v $ m = { f \over ( f + u ) } = { (f - v ) \over f } $ $ u = { -(m+1) \over m } f $ $ v = (m+1) f $ solved the equation
A convex lens of focal length f produces a real image x times the size of an object, Then what is the distance of the object fromthe lens ?
$ v = f \left( 1 -{ v\over u } \right) = f (1 -x ) $ $ \therefore u = f { (1+x ) \over x } $
A thin lens has focal length f, and its aperture has diameter d. It forms an image of intensity I. Now, the central part of the aperture upto diameter d/2 is blocked by an opeque paper. The focal length and image intensity will change to.....
Due to blocking of central part, focal lenght does not change. However, theintensity decreases. The amount of light crossing the lens decreases by a factor of $ {\pi ( { d \over 2 }) ^2 \over \pi d^2 } = { 1 \over 4 } , Hence I' = I -{I \over 4 } = { 3I \over 4 } $
The distance between object and the screen is D. Real images of an object are formed on the screen two positions of a lens seperated by a distance d. What will be the ratio between the sizes of two images ?
$ m_1 = { I _1 \over 0 } = { v \over u } = { D+d \over D-d} $ $ m_2 = { I _2 \over 0 } = { v \over u } = { D+d \over D-d} $ $ \therefore {I_1 \over I_2 } = {(D+d)^2 \over (D-d)^2 } or { I_2 \over I_1} = { (D-d)^2 \over (D+d)^2 }$
A spherical mirror forms an erect image three times the linear size of the object. If the distance between the object and the image is 80 cm, What is the focal length of the mirror ?
Knowledge base
A short linear object of length L lies on the axis of a spherical mirror of focal length of fat a distance u from the mirror. Its image has an axial length L` equal to ................
for a spherical mirrer $ v ={ uf \over u-f } …(1) $ The value of u for the two ends of the object of length are $ u _1 = \left( u - {L \over 2 } \right) and u_2 = \left(u + {L \over 2 } \right) ....(2) $ solved the eqn (1) and (2)
A concave mirror of focal length f produces an images n times the size of the object. If the image is real then What is the distance of the object from the mirror ?
$ u = ({ n+1 \over n } ) f $
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