A screen is placed at a distance of 90 cm from an object. A convex lens forms the image of this object on the screen, when placed at two different locations separated by 30 cm as shown in the figure. Determine the focal length of the lens.

The distance between the object and the screen is D = 90 cm . The lens is placed at two positions at d = 30 cm apart from each other such that the image is formed at the screen. So, the focal length of the convex lens will be: f = D 2 - d 2 4 D By substituting the values we get: f = 90 cm 2 - 30 cm 2 4 90 cm = 8100 cm 2 - 900 cm 2 4 90 cm = 7200 cm 2 360 cm = 20 cm Hence, the focal length of the lens is 20 cm .

A point source O is placed at a distance of 12 . 5 cm from a converging lens of focal length 10 cm , on its principal axis as shown in the figure. Find the distance at which a diverging mirror of focal length 20 cm should be placed, so that a real image is formed on the source itself.

Here, the distance between converging lens and point source is, u = 12 . 5 cm Focal length of the converging lens is, f = 10 cm Assuming the image I 1 of the point source is created at v . Then, using lens formula: 1 v + 1 u = 1 f and applying Cartesian sign convention. By, substituting values we get: 1 v = 1 10 cm - 1 12 . 5 cm ⇒ v = 125 cm 2 2 . 5 cm = 50 cm Since, the image is formed to the left of the lens so it will be negative. Now the convex mirror of focal length f ' = 20 cm According to the question the convex mirror is to be placed to the left of lens in such a way that light rays fall on the mirror normally. Then only the light rays will retrace their path and form final image on the object itself. So, the image I 1 formed by the lens should be at the centre of curvature of the mirror M , i.e. M I = R = 2 f ' = 40 cm The distance between the lens and the image formed through it is, I L = 50 cm . So, the distance between convex lens and convex mirror will be: M L = L I - M I = 50 cm - 40 cm = 10 cm Hence, the given diverging mirror is placed 10 cm to the left of the lens so that the real image is formed on the source itself.

With the help of a neat ray diagram, derive the relationship between the focal length, the object distance and the image distance for a concave mirror. A concave mirror is first kept in a medium of refractive index 1 . 33 and then in another medium of refractive index 1 . 4 . What will be the change in its focal length?

Here, the mirror M 1 M 2 is having pole P , focus at F (then focal length is P F = f ) and centre of curvature at C . The object is place at B , then the distance between position of object and Pole is u . The light ray coming through the object from A travel parallelly to D , and reflect in the direction of D A ' passing through the focus of the concave mirror. Another ray from A travels to P and reflects in the direction of the P A ' . The point where these two rays intersect that will be A ' and the perpendicular dropped from A ' on the principal axis that is B ' will be the position of the image so its distance from pole is v . Comparing △ A B C and △ A ' B ' C : ∠ A B C = ∠ A ' B ' C = 90 ° and ∠ A C B = ∠ A ' C B ' So, A B A ' B ' = B C B ' C .................................. 1 Comparing △ D N F and △ A ' B ' F : ∠ D N F = ∠ A ' B ' F = 90 ° and ∠ D F N = ∠ A ' F B ' So, D N A ' B ' = F N F B ' . Since, A B = D N then A B A ' B ' = F N F B ' .................................. 2 From 1 and 2 , we get: B C B ' C = F N F B ' .................................. 3 Now, assuming the aperture of the mirror to be very small then N is near to P . Then F N = F P Then equation 3 will be: P B - P C P C - P B ' = F P P B ' - P F .................................. 4 So, - u - ( - 2 f ) - 2 f - ( - v ) = - f - v - ( - f ) ⇒ - u + 2 f 2 f + v = - f - v + f 2 f 2 - v f = - u f + u v - 2 v f + 2 f 2 u f + v f = u v Now dividing both sides by u v f , we get: u f + v f u v f = u v u v f ⇒ 1 v + 1 u = 1 f Hence, the required relation for the concave mirror is 1 v + 1 u = 1 f . The focal length of the concave mirror is independent of the medium in which it is held. Hence, there will not be any change in its focal length due to change in refractive index of the medium.

A spherical surface of radius of curvature separates a rarer and a denser medium. With the help of a neat ray diagram obtain the relationship between the object distance ‘ u ’ , the image distance ‘ v ’ , the radius of curvature ‘ R ’ and the refractive indices n 1 and n 2 of the two media, when the object is kept in the rarer medium. How does the scattering of light vary with the wavelength of different colours of light ?

Assuming two mediums of different refractive index (such that n 1 > n 2 ) that has been separated by a spherical surface of radius of curvature. Here, O is the position of the object and its image is forming on I . Using Snell's law at point P : n 1 sin θ 1 = n 2 sin θ 2 ............................ 1 Since, θ 1 and θ 2 are very small, so θ 1 ≈ sin θ 1 and θ 2 ≈ sin θ 2 . So, the equation 1 will be: n 1 θ 1 = n 2 θ 2 ............................ 2 Here, θ 1 = α + β and β = θ 2 + γ ⇒ θ 2 = β - γ By substituting values of θ 1 and θ 2 in equation 2 , we get: n 1 α + β = n 2 β - γ ............................ 3 ⇒ β = n 1 n 2 α + β + γ ⇒ n 2 - n 1 β = n 1 α + n 2 β ............................ 4 Here, β = a r c P M M C = S R , α = a r c P M O M = S u and α = a r c P M M I = S v , substituting this in equation 4 , we get: (Applying Cartesian sign convention) n 2 - n 1 S R = n 1 - S u + n 2 S v n 2 - n 1 1 R = - n 1 1 u + n 2 1 v - n 1 u + n 2 v = n 2 - n 1 1 R . Hence, the required relation has been obtained that is, - n 1 u + n 2 v = n 2 - n 1 1 R Since, Rayleigh scattering is inversely proportional to the fourth power of wavelength, so the shorter wavelength (i.e. violet and blue light) will scatter more than the longer wavelengths (i.e. orange and red light).

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>Goa Board-2019>Q 6

MEDIUM

12th Goa Board

IMPORTANT

Earn 100

The radius of curvature of either face of a convex lens is equal to its focal length. Calculate the refractive index of the material of the lens.

Important Questions on Ray Optics and Optical Instruments

HARD

12th Goa Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>Goa Board-2019>Q 7

With the help of a diagram, obtain an expression for the refractive index of the material of the prism. If

θ

is the angle of incidence and the refracted ray inside the prism is parallel to the base, then what will be the angle of emergence ?

HARD

12th Goa Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>Goa Board-2019>Q 8

With the help of a diagram, obtain a relation between object distance, image distance and focal length of a concave mirror. If the concave mirror is immersed in water, will the above equation be still valid ?

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