Request a call back

Join NOW to get access to exclusive study material for best results

NEET Class neet Answered

A WATER TANK HAS A SMALL SPHERICAL WINDOW OF RADIUS OF CURVATURE 15CM..THE CONVEX SURFACE OF WINDOW IS OUTSIDE OF TANK . A POINT OBJECT IS PLACED AT A DISTANCE OF 60CMS FROM THE WINDOW IN AIR.FIND THE IMAGE DISTANCE

Asked by neetaspirantdoctor | 14 May, 2022, 10:59: PM
Expert Answer
First figure shows the refraction taking place when light rays from object O incident on front surface of glass window.
Light rays enters from air ( n1 = 1 )  to glass ( n2 = 1.5 ), where n represents refractive index
 
Image distance is determined from the following equation for refraction at spherical surace .
 
begin mathsize 14px style n subscript 2 over v minus n subscript 1 over u space equals space fraction numerator n subscript 2 minus n subscript 1 over denominator R end fraction end style ................................(1)
where v = image distance = AI1
u = object distance = AO = -60 cm
R = radius of curvature = 15 cm
 
begin mathsize 14px style fraction numerator 1.5 over denominator A I subscript 1 end fraction minus fraction numerator 1 over denominator left parenthesis negative 60 right parenthesis end fraction space equals space fraction numerator 1.5 space minus space 1 over denominator 15 end fraction end style
 
From above expression , we get AI1 = 90 cm
 
Now first image I1 is acting as object for the spherical surface of window which is inside water.
 
Light rays enters from water ( n2 = 1.333 )  to glass ( n1 = 1.5 ), where n represents refractive index
 
Image distance is determined from eqn.(1) by using proper sign convention
 
begin mathsize 14px style fraction numerator 1.5 over denominator B I subscript 2 end fraction minus fraction numerator 1.333 over denominator B I subscript 1 end fraction space equals space fraction numerator 1.333 space minus space 1.5 over denominator negative 15 end fraction end style
 
begin mathsize 14px style fraction numerator 3 over denominator 2 space B I subscript 2 end fraction minus fraction numerator 4 over denominator left parenthesis 3 space cross times space 60 right parenthesis end fraction space equals space 1 over 90 end style
 
We get final image distance BI2 = 45 cm from above expression
Answered by Thiyagarajan K | 15 May, 2022, 08:54: AM
NEET neet - Physics
Asked by myindiaisbad | 06 Jul, 2022, 03:36: PM
ANSWERED BY EXPERT
NEET neet - Physics
Asked by sirib942254 | 15 Jul, 2021, 05:32: PM
ANSWERED BY EXPERT
NEET neet - Physics
Asked by tasminmazumdar | 01 Jun, 2021, 12:48: AM
ANSWERED BY EXPERT
NEET neet - Physics
Asked by jhajuhi19 | 25 Aug, 2020, 04:40: PM
ANSWERED BY EXPERT
NEET neet - Physics
Asked by inbasri224 | 01 Feb, 2019, 11:18: AM
ANSWERED BY EXPERT
NEET neet - Physics
Asked by deepakudgiri29 | 09 Jan, 2019, 06:29: PM
ANSWERED BY EXPERT
×