Click to Chat
1800-2000-838
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Solved Examples on Ray Optics Question 1:- A praying mantis preys along the central axis of a thin symmetrical lens, 20 cm from the lens. The lateral magnification of the mantis provided by the lens is m = -0.25, and the index of refraction of the lens material is 1.65. (a) Determine the type of image produced by the lens; whether the object (mantis) is inside or outside the focal point; on which side of the lens the image appears; and whether the image is inverted. (b) What is the magnitude r of the two radii of curvature of the lens? Solution:- (a) From equation m = – i/p, and the given value for m, we see that i = -mp = 0.25p Even without finishing the calculation, we can answer the questions. Because p is positive, i here must be positive. That means we have a real image, which means we have a converging lens (the only lensthat can produce a real image). The object must be outside the focal point (the only way a real image can be produced). And the image is inverted and on the side of the lens opposite the object. (That is how a converging lens makes a real image.) (b) Because the lens is symmetric, r_{1} (for the surface nearer the object) and r_{2} have the same magnitude r. Because the lens is a converging lens, r_{1} = +r and r2 = -r. We have only one question 1/f = (n-1) [1/r_{1} – 1/r_{2}] that includes the radius of curvature of a lens, but we lack a value for f to insert in that equation. We can get f from equation 1/f = 1/p + 1/i if we first find i. So, we must finish the calculation for i, inserting the given value of p and obtaining i = (0.25) (20 cm) = 5.0 cm Now equation 1/f = 1/p + 1/i give us, 1/f = 1/p + 1/i = [{1/(20 cm)} + {1/(5.0 cm)}] from which we find, f = 4.0 cm. Equation 1/f = (n-1) [1/r_{1} – 1/r_{2}] gives us, 1/f = (n-1) [1/r_{1} – 1/r_{2}] = (n-1) [1/(+r) – 1/(-r)] or, with known values inserted, 1/(4 cm) = (1.65 – 1) (2/r), which gives, r = (2) (0.65) (4.0 cm0 = 5.2 cm From the above observation we conclude that, the magnitude r of the two radii of curvature of the lens would be 5.2 cm. ____________________________________________________________________________________________ Question 2 (IIT-JEE):- A small object is placed 45 cm from a convex refracting surface of radius of curvature 15 cm. If the surface separates air from glass of refractive index 1.5, find the position of the image. Also, determine the first and second principal focal lengths. Solution:- Distance of object, u = -45 cm Radius of curvature, R = +15 cm μ_{2} = 1.5 μ_{2} = 1 As the object lies in the rarer medium μ_{2}/v – μ_{1}/u = μ_{2} – μ_{1}/ R [1.5/v] – [1/(-45)] = 0.5/15 = 1/30 1.5/v = 1/30 – 1/45 1.5/v = 1/90 v = 135 cm Therefore position of the image would be at 135 cm. First principal focal length is given by f_{1} = - μ_{1}R/[μ_{2}–μ_{1}] = – (1×15)/0.5 = -30 cm Second principal focal length is given by f_{2} = - μ_{2}R/[μ_{2}–μ_{1}] = – (1.5×15)/0.5 = -45 cm From the above observation we conclude that, the first focal length would be at distance -30 cm and second focal length would be at distance -45 cm. _________________________________________________________________________________________ Question 3:- Complete AIPMT/AIIMS Course and Test Series OFFERED PRICE: R 15,000 View Details Get extra R 3,000 off USE CODE: CART20
A praying mantis preys along the central axis of a thin symmetrical lens, 20 cm from the lens. The lateral magnification of the mantis provided by the lens is m = -0.25, and the index of refraction of the lens material is 1.65.
(a) Determine the type of image produced by the lens; whether the object (mantis) is inside or outside the focal point; on which side of the lens the image appears; and whether the image is inverted.
(b) What is the magnitude r of the two radii of curvature of the lens?
(a) From equation m = – i/p, and the given value for m, we see that
i = -mp
= 0.25p
Even without finishing the calculation, we can answer the questions. Because p is positive, i here must be positive. That means we have a real image, which means we have a converging lens (the only lensthat can produce a real image). The object must be outside the focal point (the only way a real image can be produced). And the image is inverted and on the side of the lens opposite the object. (That is how a converging lens makes a real image.)
(b) Because the lens is symmetric, r_{1} (for the surface nearer the object) and r_{2} have the same magnitude r. Because the lens is a converging lens, r_{1} = +r and r2 = -r. We have only one question 1/f = (n-1) [1/r_{1} – 1/r_{2}] that includes the radius of curvature of a lens, but we lack a value for f to insert in that equation. We can get f from equation 1/f = 1/p + 1/i if we first find i. So, we must finish the calculation for i, inserting the given value of p and obtaining
i = (0.25) (20 cm)
= 5.0 cm
Now equation 1/f = 1/p + 1/i give us,
1/f = 1/p + 1/i
= [{1/(20 cm)} + {1/(5.0 cm)}]
from which we find, f = 4.0 cm.
Equation 1/f = (n-1) [1/r_{1} – 1/r_{2}] gives us,
1/f = (n-1) [1/r_{1} – 1/r_{2}]
= (n-1) [1/(+r) – 1/(-r)]
or, with known values inserted,
1/(4 cm) = (1.65 – 1) (2/r),
which gives,
r = (2) (0.65) (4.0 cm0
= 5.2 cm
From the above observation we conclude that, the magnitude r of the two radii of curvature of the lens would be 5.2 cm.
____________________________________________________________________________________________
A small object is placed 45 cm from a convex refracting surface of radius of curvature 15 cm. If the surface separates air from glass of refractive index 1.5, find the position of the image. Also, determine the first and second principal focal lengths.
Distance of object, u = -45 cm
Radius of curvature, R = +15 cm
μ_{2} = 1.5
μ_{2} = 1
As the object lies in the rarer medium
μ_{2}/v – μ_{1}/u = μ_{2} – μ_{1}/ R
[1.5/v] – [1/(-45)] = 0.5/15
= 1/30
1.5/v = 1/30 – 1/45
1.5/v = 1/90
v = 135 cm
Therefore position of the image would be at 135 cm.
First principal focal length is given by
f_{1} = - μ_{1}R/[μ_{2}–μ_{1}]
= – (1×15)/0.5
= -30 cm
Second principal focal length is given by
f_{2} = - μ_{2}R/[μ_{2}–μ_{1}]
= – (1.5×15)/0.5
= -45 cm
From the above observation we conclude that, the first focal length would be at distance -30 cm and second focal length would be at distance -45 cm.
_________________________________________________________________________________________
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.