An insect that is 5.00 mm long is placed \(10.0 \mathrm{cm}\) from a converging lens with a focal length of \(12.0 \mathrm{cm} .\) (a) What is the position of the image? (b) What is the size of the image? (c) Is the image upright or inverted? (d) Is the image real or virtual? (e) What is the angular magnification if the lens is close to the eye?

#tag_title# Step 3: Calculate the image distance (d_i) #tag_content#Using the lens formula, we can find the image distance: \(\frac{1}{d_i} = \frac{1}{12} - \frac{1}{10}\) Now, we calculate the common denominator and subtract the fractions: \(\frac{1}{d_i} = \frac{10 - 12}{120}\) \(\frac{1}{d_i} = -\frac{2}{120}\) Now, we take the reciprocal to find d_i: \(d_i = -60\) cm The negative sign indicates that the image is on the same side of the lens as the object, which means it is a virtual image. #tag_title# Step 4: Calculate the magnification #tag_content#Magnification (m) is given by the ratio of the image distance to the object distance: \(m = \frac{d_i}{d_o}\) Using the given values, we can calculate the magnification: \(m = \frac{-60}{10}\) \(m = -6\) The negative magnification indicates that the image is inverted relative to the object. #tag_title# Step 5: Calculate the image height #tag_content#Now, we can find the image height (h_i) using the magnification: \(h_i = m \times h_o\) We already know the magnification and object height: \(h_i = -6 \times 5.00\) mm \(h_i = -30.0\) mm The image is 30.0 mm tall and inverted. #tag_title# Step 6: Determine the properties of the image #tag_content#The image properties can be summarized as follows: - Position: The image is 60 cm from the lens on the same side as the object (virtual image). - Magnification: The image is 6 times larger than the object (magnification of -6). - Orientation: The image is inverted, as indicated by the negative magnification. - Size: The image height is 30.0 mm. #tag_title# Step 7: Calculate the angular magnification #tag_content#Angular magnification (M) is given by the ratio of the image size to the object size: \(M = -m\) Using the magnification we obtained earlier: \(M = -(-6)\) \(M = 6\) Therefore, the angular magnification is 6. #tag_title# Short Answer #tag_content#The image of the insect is 60 cm away on the same side of the lens as the object, forming a virtual image. The image is 30.0 mm tall, which is 6 times larger than the object, and it is inverted. The angular magnification is 6.