Question

An image Y is formed of a point X by a Mirror whose principal axis AB is shown .Draw a ray diagram to locate the focus and the Mirror.

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Answer 1

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GɪᴠᴇN:-

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

Tᴏ FɪɴD:-

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

AɴsᴡᴇR :-

$\underline{\underline{\textsf{\textbf{\purple{\red{ \leadsto }\: Diagram:}}}}}$

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

( 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.

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

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

• 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.

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Answer 2

Answer:

ok

Explanation: