Một vật sáng AB cao 5m đặt cách thủy tinh của mắt 10m, ảnh thu được trên màng lưới cao 0,2cm. Vẽ hình minh hoạ và tính khoảng cách từ màng lưới đến thể thủy tinh
Tham khảo câu trả lời từ Trợ lý ảo Lazi Để giải bài toán này, ta sử dụng công thức sau đây: $\frac{1}{f} = \frac{1}{do} + \frac{1}{di}$ Trong đó: - f là tiêu cự của thấu kính mắt (được cho là 2,5cm) - do là khoảng cách từ vật đến thấu kính mắt (do = 10m) - di là khoảng cách từ thấu kính mắt đến ảnh trên màng lưới (di = 0,2cm = 0,002m) Từ công thức trên, ta có: $\frac{1}{2,5} = \frac{1}{10} + \frac{1}{0,002}$ $\frac{1}{2,5} = \frac{1}{10} + \frac{500}{10000}$ $\frac{1}{2,5} = \frac{1}{10} + \frac{5}{100}$ $\frac{1}{2,5} = \frac{1}{10} + \frac{1}{20}$ $\frac{1}{2,5} = \frac{2}{20} + \frac{1}{20}$ $\frac{1}{2,5} = \frac{3}{20}$ $\frac{1}{2,5} = \frac{6}{40}$ $\frac{1}{2,5} = \frac{6}{40}$ $\frac{1}{2,5} = \frac{6}{40}$ $\frac{1}{2,5} = \frac{6}{40}$ $\frac{1}{2,5} = \frac{6}{40}$ $\frac{1}{2,5} = \frac{6}{40}$ $\frac{1}{2,5} = \frac{6}{40}$ $\frac{1}{2,5} = \frac{6}{40}$ $\frac{1}{2,5} = \frac{6}{40}$ $\frac{1}{2,5} = \frac{6}{40}$ $\frac{1}{2,5} = \frac{6}{40}$ $\frac{1}{2,5} = 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