ITEM 2 a) Find the nature, size, and position of the image of the boy on the camera. Step 1: Determine the image position ($v$) using the lens formula. Given: Focal length $f = 40 \text{ cm}$ (converging lens, so positive), object distance $u = 160 \text{ cm}$. The lens formula is: $$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$ Substitute the given values: $$\frac{1}{40 \text{ cm}} = \frac{1}{160 \text{ cm}} + \frac{1}{v}$$ Rearrange to solve for $\frac{1}{v}$: $$\frac{1}{v} = \frac{1}{40 \text{ cm}} - \frac{1}{160 \text{ cm}}$$ Find a common denominator: $$\frac{1}{v} = \frac{4}{160 \text{ cm}} - \frac{1}{160 \text{ cm}}$$ Subtract the fractions: $$\frac{1}{v} = \frac{3}{160 \text{ cm}}$$ Invert to find $v$: $$v = \frac{160}{3} \text{ cm}$$ $$v \approx 53.33 \text{ cm}$$ The position of the image is $\boxed{\text{53.33 cm from the lens on the opposite side of the object}}$. Step 2: Determine the magnification ($M$) and image size ($h_i$). Given: Object height $h_o = 120 \text{ cm}$. The magnification formula is: $$M = -\frac{v}{u} = \frac{h_i}{h_o}$$ Substitute the values for $v$ and $u$: $$M = -\frac{160/3 \text{ cm}}{160 \text{ cm}}$$ $$M = -\frac{1}{3}$$ Now, calculate the image height $h_i$: $$h_i = M \times h_o$$ $$h_i = -\frac{1}{3} \times 120 \text{ cm}$$ $$h_i = -40 \text{ cm}$$ The size of the image is $\boxed{\text{40 cm}}$. Step 3: Determine the nature of the image. Since $v$ is positive, the image is formed on the opposite side of the lens from the object, indicating a real image. Since $M$ is negative, the image is inverted. Since $|M| = \frac{1}{3} < 1$, the image is diminished. The nature of the image is $\boxed{\text{real, inverted, and diminished}}$. b) Explain to your fellow students and the literature teacher the cause of the problem and the would-be possible solution to the above problem. The student's inability to see clearly both near objects (letters in books) and distant objects (board and teacher) suggests a severe refractive error. The most common cause for such symptoms is myopia (short-sightedness), where the eye's lens or cornea is too curved, or the eyeball is too long. This causes light to focus in front of the retina instead of directly on it, resulting in blurry vision for distant objects and potentially near objects if the condition is severe. The solution to this problem is to correct the refractive error using concave lenses (diverging lenses). These lenses diverge the light rays before they enter the eye, effectively shifting the focal point backward onto the retina, allowing the student to see clearly. The student should visit an optometrist or ophthalmologist for an eye examination to get a precise prescription for corrective eyeglasses or contact lenses. That's 2 down. 3 left today — send the next one.