how to derive mirror formula
Answers
Answered by
4
★★★hiii friend!!!!★★★
___________________________________________________________
■Here is ur questions answer!!
⇒ Mirror formula is the relationship between object distance (u), image distance (v) and focal length.
⇒ 1 upon v+1 upon u=1 upon f
■ Now, we have to proof of mirror formula:
⇒ Let in ΔABCΔABC and ΔA′B′C′
√ ΔABCΔABC ∼∼ ΔA′B′CΔA′B′C [AA similarity]
√ AB upon A′B′=AC upon A′C′ ⋯⋯⋯⋯⋯⋯⋯⋯(1)
√ Similarly, In ΔFPEΔFPEand ΔA′B′F′
√ EP upon A′B′=PF upon A′F
√ AB upon A′B′=PF upon A′F [ AB=EP] ⋯⋯(II)
⇒ now From (i) &(ii)
√ AC upon A′C=PF upon A′F
=>A′C upon AC=A′F upon PF
=>(CP−A′P) upon (AP−CP)=(A′P–PF) upon /PF
→ Now, PF=−f;CP=2PF=−2fPF=−f;CP=2PF=−2f ; AP=−uAP=−u ; and A′P=−v
√ Put these value in above relation:
⟹[(−2f)–(−v)] upon( −u)−(−2f) =[(−v)–(−f)] upon (−f)
⟹uv=fv+uf
⟹1upon f =1 upon u + 1 upon v
___________________________________________________________
★★★hope this may help u!!!★★★
___________________________________________________________
■Here is ur questions answer!!
⇒ Mirror formula is the relationship between object distance (u), image distance (v) and focal length.
⇒ 1 upon v+1 upon u=1 upon f
■ Now, we have to proof of mirror formula:
⇒ Let in ΔABCΔABC and ΔA′B′C′
√ ΔABCΔABC ∼∼ ΔA′B′CΔA′B′C [AA similarity]
√ AB upon A′B′=AC upon A′C′ ⋯⋯⋯⋯⋯⋯⋯⋯(1)
√ Similarly, In ΔFPEΔFPEand ΔA′B′F′
√ EP upon A′B′=PF upon A′F
√ AB upon A′B′=PF upon A′F [ AB=EP] ⋯⋯(II)
⇒ now From (i) &(ii)
√ AC upon A′C=PF upon A′F
=>A′C upon AC=A′F upon PF
=>(CP−A′P) upon (AP−CP)=(A′P–PF) upon /PF
→ Now, PF=−f;CP=2PF=−2fPF=−f;CP=2PF=−2f ; AP=−uAP=−u ; and A′P=−v
√ Put these value in above relation:
⟹[(−2f)–(−v)] upon( −u)−(−2f) =[(−v)–(−f)] upon (−f)
⟹uv=fv+uf
⟹1upon f =1 upon u + 1 upon v
___________________________________________________________
★★★hope this may help u!!!★★★
Attachments:
Answered by
3
Answer:
Here is your answer
Explanation:
U can see the shaded triangle which are congruent
Thank u
Attachments:
Similar questions