Đáp án: Đặt vật AB = 18 cm có hình mũi tên trước một thấu kính (AB vuông góc với trục chính và A thuộc trục chính) Ảnh AʹBʹ cách thấu kính cùng chiều với AB và có độ cao bằng 1/3AB a) Thấu kính này là thấu kính gì? Vì sao b) Ảnh AʹBʹ cách thấu kính 9cm Vẽ hình và tín... - a, Ảnh cùng chiều, nhỏ hơn vật ⇒ Đây là thấu kính phân kì b, Ta có ΔABO∼ΔAʹBʹO ⇒(AB/AʹBʹ)=(AO/AʹO) ⇔AO=(ABAʹO/AʹBʹ)=3AʹO=39=27cm Ta có ΔIOF∼ΔBʹAʹF ⇒(IO/AʹBʹ)=(OF/AʹF) ⇔3=(OF/OF-AʹO) ⇔3=(OF/OF-9) ⇔OF=13,5cm

a, Ảnh cùng chiều, nhỏ hơn vật ⇒ Đây là thấu kính phân kì b, Ta có ΔABO∼ΔAʹBʹO ⇒(AB/AʹBʹ)=(AO/AʹO) ⇔AO=(ABAʹO/AʹBʹ)=3AʹO=39=27cm Ta có ΔIOF∼ΔBʹAʹF ⇒(IO/AʹBʹ)=(OF/AʹF) ⇔3=(OF/OF-AʹO) ⇔3=(OF/OF-9) ⇔OF=13,5cm

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2 năm trước

Đặt vật AB = 18 cm có hình mũi tên trước một thấu kính (AB vuông góc với trục chính và A thuộc trục chính). Ảnh A'B' cách thấu kính cùng chiều với AB và có độ cao bằng 1/3AB a) Thấu kính này là thấu kính gì? Vì sao b) Ảnh A'B' cách thấu kính 9cm. Vẽ hình và tính tiêu cự

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Trương Tiến Nhật

2 năm trước

a, Ảnh cùng chiều, nhỏ hơn vật

$⇒$ Đây là thấu kính phân kì

b, Ta có: $ΔABO\simΔA'B'O$

$⇒\dfrac{AB}{A'B'}=\dfrac{AO}{A'O}$

$⇔AO=\dfrac{AB.A'O}{A'B'}=3.A'O=3.9=27cm$

Ta có: $ΔIOF\simΔB'A'F$

$⇒\dfrac{IO}{A'B'}=\dfrac{OF}{A'F}$

$⇔3=\dfrac{OF}{OF-A'O}$

$⇔3=\dfrac{OF}{OF-9}$

$⇔OF=13,5cm$

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