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
What are the sign conventions followed for lenses?
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.
Derive the equation for thin lens and obtain its magnification.
Derive the equation for effective focal length for lenses in out of contact.
If the distance D between an object and screen is greater than 4 times the focal length of a convex lens, then there are two positions of the lens for which images are formed on the screen. This method is called conjugate foci method. If d is the distance between the two positions of the lens, obtain the equation for a focal length of the convex lens.
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.
