Using lens maker’s formula, derive the thin lens formula 1 f = 1 v - 1 u for a biconvex lens.

The lens maker's formula is 1 v - 1 u = n - 1 1 R 1 - 1 R 2 ....................... 1 Here, n = n 2 n 1 that is the ratio of refractive index of lens to other medium. For a thin biconvex lens, assuming the object is at infinity (i.e. u = ∞ ) and the image is forming at focal point (i.e. v = f ). Substituting the above values in lens maker's formula we get, 1 f = n - 1 1 R 1 - 1 R 2 ..................... 2 From 1 and 2 , we get 1 v - 1 u = 1 f . Hence, thin lens formula 1 f = 1 v - 1 u has been derived for a biconvex lens.

Using lens maker’s formula, derive the thin lens formula 1 f = 1 v - 1 u for a biconvex lens.

The lens maker's formula is 1 v - 1 u = n - 1 1 R 1 - 1 R 2 ....................... 1 Here, n = n 2 n 1 that is the ratio of refractive index of lens to other medium. For a thin biconvex lens, assuming the object is at infinity (i.e. u = ∞ ) and the image is forming at focal point (i.e. v = f ). Substituting the above values in lens maker's formula we get, 1 f = n - 1 1 R 1 - 1 R 2 ..................... 2 From 1 and 2 , we get 1 v - 1 u = 1 f . Hence, thin lens formula 1 f = 1 v - 1 u has been derived for a biconvex lens.

Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position. A giant refracting telescope at an observatory has an objective lens of focal length 15 m and an eyepiece of focal length 1 . 0 cm . If this telescope is used to view the Moon, find the diameter of the image of the Moon formed by the objective lens. The diameter of the Moon is 3 . 48 × 10 6 m , and the radius of the lunar orbit is 3 . 8 × 10 8 m .

Astronomical telescope in near point adjustment : Given : Focal length if objective lens, f o = 15 m Focal length of eye piece, f e = 1 cm = 0 . 1 m Diameter of the moon, d = 3 . 48 × 10 6 m Radius of lunar orbit, u o = 3 . 8 × 10 8 m Size of the image of the moon, I = ? The angle subtended by the moon at the objective lens, α = d u o — equation i And the angle subtended by the image of the moon at objective lens is, β = I f o — equation ii Both these angles in equations i and ii are equal. d u o = I f o ∴ I = d × f o u o Substitute all the values from above I = 3 . 48 × 10 6 × 15 3 . 8 × 10 8 = 13 . 73 cm

A symmetric biconvex lens of the radius of curvature R and made of glass of refractive index 1 . 5 , is placed on a layer of liquid placed on top of a plane mirror as shown in the figure. An optical needle with its tip on the principal axis of the lens is moved along the axis until its real, inverted image coincides with the needle itself. The distance of the needle from the lens is measured to be x . On removing the liquid layer and repeating the experiment, the distance is found to be y . Obtain the expression for the refractive index of the liquid in terms of x and y .

We are given, the refractive index of the lens, μ g = 1 . 5 . The distance of the needle from the lens in the first case is equal to the focal length of the combination of a convex lens and a planoconcave lens formed by the liquid. Thus, f = x The distance measured in the second case is equal to the focal length of the convex lens. Thus, f 1 = y Let f 2 be the focal length of the planoconcave lens formed by the liquid. Then, 1 f = 1 f 1 + 1 f 2 1 f 2 = 1 x - 1 y = y - x x y f 2 = x y y - x Let the radius of curvature of the first surface be R The radius of curvature of the second surface will be - R Thus, for the biconvex lens, 1 f 1 = μ g - 1 1 R - 1 - R 1 y = 1 . 5 - 1 2 R ⇒ y = R Let the refractive index of liquid be μ . For plano-convex lens, R 2 = ∞ R 1 = - R Thus, 1 f 2 = μ - 1 1 - R - 1 ∞ Substituting values from above, y - x x y = μ - 1 1 - y μ - 1 = - ( y - x ) x μ = x - y x + 1 μ = 2 x - y x

The figure shows a ray of light falling normally on the face AB of an equilateral glass prism having a refractive index 3 / 2 , placed in water of refractive index 4 / 3 . Will this ray suffer total internal reflection on striking the face AC? Justify your answer.

It is given that the glass prism is equilateral. μ p = 3 / 2 μ w = 4 / 3 Thus, ∠ A = 60 ° . Let us draw a normal to the face AC. From the geometry, we can see, the angle of incidence on the face AC is 60 ° . Thus, i = 60 ° . Let this angle be the critical angle i.e. i = i c . Then the refractive index of the prism in the air, μ = 1 sin i c = 1 sin 60 ° μ = 2 3 = 1 . 54 Now if the prism is placed in water then the refractive index of a prism with water, μ A = μ p μ w = 3 2 4 3 = 9 8 = 1 . 125 The total internal reflection at face AC, is possible only when μ < μ A . But from the results obtained above, we can see that μ > μ A ∴ 1 . 54 > 1 . 125 No total internal reflection takes place and the light ray gets refracted from the prism placed in water. Thus, we can conclude that for total internal reflection, in this case, the Critical angle should be greater than 60 ° .

Three lenses of focal length + 10 c m , - 10 c m , + 30 c m are arranged coaxially as in the figure given below. Find the position of the final image formed by the combination.

Given: Focal length of lens 1 = + 10 c m Focal length of lens 2 = – 10 c m Focal length of lens 3 = + 30 c m Applying lens formula, we get 1 f = 1 v - 1 u --------- (1) For lens 1, we have f = + 10 , v = x , and u = – 30 c m . Therefore on putting values we get, x = 15 c m Thus, for lens 2, the image of lens 1 at distance x = 15 c m is the object. Therefore, f = - 10 c m , v = x , u = 15 c m - 5 c m = 10 c m Again, putting values in equation (1) : v = ∞ Thus, for lens 3, image of lens 2 is the object, which is located at infinity. We know that for a convex lens, rays coming from infinity converge at the focus. Thus, image is formed at the focus of lens 3, that is, at 30 c m . Thus, the final image formed by the combination is at 30 c m .

Under what conditions is the phenomenon of total internal reflection of light observed? Obtain the relation between the critical angle of incidence and the refractive index of the medium.

Total internal reflection is defined as the phenomenon of reflection of light into a denser medium from an interface of this denser medium and a rarer medium. The conditions for total internal reflection: (i) Incident light ray should travel from a denser medium to a rarer medium. (ii) Angle of incidence, i, should be greater than the critical angle, C, for pair of media in contact. Critical angle is defined as the angle of incidence in denser medium to which angle of refraction is 90° in the rarer medium. Relation between the critical angle of incidence and refraction index of the medium. Consider angle of incidence is equal to the critical angle, that is, i = c Then, the angle of refraction is r = 90°. Applying Snell’s law, we have μ d sin C = μ r sin r ,where μ d is refractive index of denser medium and μ r is refractive index of rarer medium. Therefore, μ = μ d μ r = sin r sin C = sin 90 sin C = 1 sin C sin C = 1 μ This is the required relation between critical angle of incidence and refractive index of medium.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>cbse-2020>Q 14

HARD

12th CBSE

IMPORTANT

Earn 100

Using lens maker’s formula, derive the thin lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ for a biconvex lens.

Important Questions on Ray Optics and Optical Instruments

MEDIUM

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>cbse-2020>Q 15

Draw the ray diagram of an astronomical telescope when the final image is formed at infinity. Write the expression for the resolving power of the telescope.

EASY

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>cbse-2020>Q 16

An astronomical telescope has an objective lens of focal length $20 m$ and an eyepiece of focal length $1 cm$ . Find the angular magnification of the telescope.

MEDIUM

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>cbse-2020>Q 17

An astronomical telescope has an objective lens of focal length

20 m

and an eyepiece of the focal length of

1 cm

. If this telescope is used to view the Moon, find the diameter of the image formed by the objective lens. Given the diameter of the Moon is

3.5 \times 10^{6}

m and the radius of the lunar orbit is

3.8 \times 10^{8}

m.

MEDIUM

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>cbse-2020>Q 18

An object is placed in front of a concave mirror. It is observed that a virtual image is formed. Draw the ray diagram to show the image formation and hence derive the mirror equation