Question

08 Alens forms an image at Y of an object at x Where does the lens form an image of an​

Answer 1

Complete QuestioN :-

• An image Y is formed of a point X by a Mirror whose principal axis AB is shown

GiveN :-

• An image Y is formed of a point X by a Mirror whose principal axis AB is shown

To FinD :-

• To draw a ray diagram and locate the Mirror and the focus .

AnsweR :-

D I A G R A M :-

$\setlength{\unitlength}{1 cm}\begin{picture}(12,12)\qbezier(0,0)( - 2, 2.5)(0,5)\put(0.01,5){\line(-1,-1){0.6}}\put( - 0.4,4.4){\line(-1,-1){0.6}}\put( - 0.7,3.8){\line(-1,-1){0.4}}\put( - 0.9,3.3){\line(-1,-1){0.3}}\put( - 0.9,3.2){\line(-1,-1){0.36}}\put( - 1,2.69){\line(-1,-1){0.36}}\put( - 1,2.2){\line(-1,-1){0.36}}\put( - 0.9,1.8){\line(-1,-1){0.36}}\put( - 0.8,1.4){\line(-1,-1){0.36}}\put( - 0.65,1){\line(-1,-1){0.36}}\put( - 0.4,0.6){\line(-1,-1){0.36}}\put( - 0.2,0.2){\line(-1,-1){0.36}}\put( - 2,2.4){\line(1,0){6}}\put( - 1,2.4){\line(2, - 1){3}}\put(2,0.9){\line(0,1){3}}\put(2,3.9){\line( - 2, - 1){3}}\put(2.01,0.9){\line( - 1,2){2.1}}\put(2.1,2.1){ \sf I }\put(4,2.1){ \sf B }\put(1.1,2.1){ \sf C }\put( - 2,2.1){ \sf A }\put( - 1.4,2.6){ \sf P }\put( 0.1,2.6){ \sf F }\put(2.1,4){ \sf N }\put(0.8,3.5){ \sf X }\put(0,1.9){\vector(2, - 1){1}}\put(1.272,2.36){\vector( - 1, 2){1}}\put(1.272,2.3601){\vector( 1, - 2){0.5}}\put(1,3.4){\vector( - 2,-1){1}}\put(5,-2.5,){ \boxed{\sf @RishabhRanjan} }\end{picture}$

( May refer to attachment also )

$\underline{\pink{\sf Steps\:of\: construction:-}}$

• From Y draw perpendicular on the principal axis AB such that YI = IN .

• Draw a line joining points in and X so that it meets the principal axis AB at P . The point P will be the pole of the Mirror.

• As the image why of object X is real inverted and enlarged the mirror must within Concave .

• Join Y to X and extend it towards the mirror. It represent a light ray which after striking the mirror is reflected along the same path . Therefore the point C where YX intesects the axis is a centre of curvature of the Mirrror.

• Taking C as a centre and CP is radius draw the arc of the circle . This Arc represent the concave mirror.

•The midpoint of CP is the focus F , since 2f = R.

$\rule{200}2$

$\Large{\underline{\underline{\red{\sf{More\:To\:Know:- }}}}}$

$\underline{\pink{\sf Important\: points\:about\: Spherical\:Mirrors:-}}$

As an object is held in front of a spherical mirror the distance of the object is always negative.

The real image is formed in front of the mirror so it's distance is taken as negative.

The virtual image is formed at the back of the mirror so it's distance is always positive.

Focal length of concave mirror is considered as negative.

Focal length of convex mirror is considered as positive.

When image formed is virtual and erect magnification is positive.

When image formed is real and inverted magnification is negative.

The height of the object taken to be positive as the object is usually placed on the principal axis.

The height of the image should be taken as positive for virtual images and it is taken negative for real images.

$\rule{200}4$