Advertisements
Advertisements
प्रश्न
A thin converging glass lens made of glass with refractive index 1.5 has a power of + 5.0 D. When this lens is immersed in a liquid of refractive index n, it acts as a divergent lens of focal length 100 cm. What must be the value of n?
Advertisements
उत्तर
`"P"_"a"/"P"_"l" = ((mu_"g"/mu_"a" - 1))/((mu_"g"/mu_l - 1)) = 5/(- 100//100)` = - 5
`- 5 (mu_"g"/mu_l - 1) = mu_"g"/mu_"l" - 1`
`(1.5/mu_l - 1) = 1/5 (1.5 - 1)`
`mu_l = 1.5/0.9 = 5/3`
APPEARS IN
संबंधित प्रश्न
An object is placed in front of a convex mirror of focal length of f and the maximum and minimum distance of an object from the mirror such that the image formed is real and magnified.
The radius of curvature of curved surface at a thin planoconvex lens is 10 cm and the refractive index is 1.5. If the plane surface is silvered, then the focal length will be.
Arrive at lens equation from lens maker’s formula.
Obtain the equation for lateral magnification for thin lens.
What is power of a lens?
Derive the equation for effective focal length for lenses in contact.
Obtain lens maker’s formula and mention its significance.
Derive the equation for effective focal length for lenses in out of contact.
A point object is placed at 20 cm from a thin plano-convex lens of focal length 15 cm whose plane surface is silvered. Locate the position and nature of the final image.
Find the ratio of the intensities of lights with wavelengths 500 nm and 300 nm which undergo Rayleigh scattering.
