A convex lens has 20 cm focal length in air. What is the focal length in water? [Refractive index of water = 1 . 33 , Refractive index of air-glass = 1 . 5 ]

For a lens in air incident is close to one. For water it is 1 . 33 or so. For various types of glass it's often in the naighborhood of 1 . 5 . You can experience the effect yourself if you try to read some, waterproof, printing under water. Your eyes' focal length change dramatically. The focal length in water is 78 . 125 c m .

The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm , its focal length is 12 cm . What is the refractive index of glass?

I f F = + 12 c m , R 1 = 10 c m , R 2 = 15 c m , h = ? n - 1 = 0 . 5 n = 0 . 5 + 1 n = 0 . 5 + 1 = n = 1 . 5 .

If f = 0 . 5 m for a glass lens, what is the power of the lens?

Given f = 0 . 5 m For a glass lens and the power of the lens is - P o w e r = + 2 d i o p t r e .

Deduce the relation between object and image distance in terms of refractive index of the medium and the radius of curvature of curved spherical surface.

OM= V= object distance MI= V= image distance MC= R= radius of curvature Angle i= angle incidence Angle r= angle of refraction ON= incidence day N= refracted day NC= normal and n 1 , n 2 , are the refractive .

Draw a neat ray diagram to show the image formation in a compound microscope. Define its magnifying power and derive an expression for it, when the final image is at near point of eye.

The objective lens form image A ' B ' near the first focal point of eyepiece. m o = linear magnification Since m o = h ' h Aperture and focal length increase or decrease the resolving power of the compound microscope. Resolving power of microscope is given by R . P = 2 n sin θ / 1 . 22 λ (i) On decreasing the aperture (diameter) of the objective lens, value of sin q decreases, and hence resolving power decreases. (ii) On decreasing the focal length of the objective lens, value of sin q increases and hence resolving power increases.

Derive the lens maker’s formula for a thin double convex lens. Why is it called so?

We know that from the refraction formula = n 1 - u + N 2 v = n 2 - n 1 r , - n 1 - u + n 2 v = n 2 - N 1 r 1 = n 1 - u + n 2 v = n 2 - n 1 r = - n 2 - u + n 1 v = n 2 - n 1 - r 2 A d d i n g 1 a n d 2 w e g e t n 1 - u + n 1 v = n 2 - n 1 1 r 1 - 1 r 2 o r 1 - u + 1 v = n 2 - n 1 n 1 1 r 1 - 1 r 2 W h e n u = ∞ , v = f W h e n u = f , v = ∞ ∴ T h e t e r m o n t h e L H S c a n b e r e p l a c e d b y 1 f w h e r e f i s t h e f o c a l l e n g t h o f l e n s 1 f = n 2 - n 1 n 1 1 r 1 - 1 r 2 T h e e q u a t i o n ( 4 ) i s c a l l e d t h e l e n s m a k e f o r m u l a 1 N 2 = 1 - r 1 r 2 f r 1 - r 2 w h e r e 1 N 2 = n 2 n 1 ( 2 ) f o r r a d i u s o f c u r v a t u r e t h e l e t t e r R m a y b e u s e d .