Question

What type of magnification compares the angular size of an image to the angular size of an object if it were at a reference distance of 25cm?

Answer 1

The type of magnification that compares the angular size of an image to the angular size of an object at a reference distance of 25 cm is angular magnification.

Step-by-step explanation:

Angular magnification
`(\(M_a\))`) is a measure of how much larger an object appears when viewed through an optical instrument. It is defined as the ratio of the angle subtended by the image to the angle subtended by the object when both are measured at the eye. Mathematically, it is expressed as:

`\[ M_a = \frac{\theta_{\text{image}}}{\theta_{\text{object}}} \]`

where
`\(\theta_{\text{image}}\)` is the angular size of the image and
`\(\theta_{\text{object}}\)` is the angular size of the object. In this case, the reference distance is given as 25 cm. The angular size
`(\(\theta\))` is given by the formula:

`\[ \theta = \tan^(-1)\left((h)/(d)\right) \]`

where \(h\) is the height of the object or image, and
`\(d\)` is the distance from the object or image to the observer. At the reference distance, the angular magnification becomes:

`\[ M_a = \frac{\tan^(-1)\left(\frac{h_{\text{image}}}{25}\right)}{\tan^(-1)\left(\frac{h_{\text{object}}}{25}\right)} \]`

This formula quantifies how much larger the image appears compared to the object when both are viewed from a distance of 25 cm.

In conclusion, angular magnification is the appropriate measure for comparing the perceived size of an image to the actual size of an object, taking into account the angle subtended by both at a specified reference distance.