Test: Refraction through Spherical surfaces(27 Dec) - JEE MCQ

When the lens is in air, we have (from lens maker's equation), 1/20 = [(1.5/1) - 1) (1/R1- 1/R2). 
We are not bothered about the signs of R1 and R2 since they are unknown quantities. 
Even though we know that the convergent lens will become divergent in the denser medium, we write the lens maker's equation without bothering about the sign of its unknown focal length in the liquid. If if is the focal length in the liquid, we can write, 
1/f = ((1.5/1.6) - 1) (1/R1 - 1/R2). 
Dividing the first equation by the second, we obtain f/20 = 0.5x1.6/(-0.1) from which f = -160 cm

Using the lens maker’s formula we get
p= (μ−1) (1/R₁​+1/R₂​)
As we can clearly see, when R₁= R₂= 5cm then the maximum power is achieved.

Consider refraction of light from infinity at first surface,
(μ2/v)​​−( μ1/u​)​=(μ2​−μ1/R)​​
(1.5​/ v1​)-(1.75/−∞)​=1.5−1.75​/−R
Consider refraction of this image from second surface,
​(1.75​/ v2)−(​1.5/v1)​=(1.75−1.5​)/R
Hence, v2​=3.5R
This is the image of light coming from infinity, therefore the focal length.
It is converging since focal length is positive.
 

M=−ϑ /u ​=−m/1​
ϑ=−mu
(1/ϑ​)−(1/u)​=1/f
1/mu​+1/u​=−1/f​
(1+m)​/xu=−1/f​
u= (1+m)​f/(m)

Applying Snell's law at P ,



The rays make an angle of with each other .

Combined focal length



Magnification of the lens combination in normal adjustment is

Considering refraction at the curved surface,
u = −20, μ₂ = 1
μ₁ = 3/2, R = +20

i.e., 10 cm below the curved surface or 10 cm above the actual position of flower.

When a light ray passes from one medium to another medium let's say from air to prism, the frequency (f) of the light ray remains the same but the velocity (v) and wavelength (λ) of the light ray changes because the frequency of the light wave is decided only by the source whereas the velocity and wavelength of a wave depends on the nature of the medium in which travels.
For a wave, the relation between the velocity, wavelength and frequency is v = λf.