An object is located 25cm in front of a +6.00 lens, where is the image? is it real or virtual?
An obj ect is located 25 cm in front of a + 6.pp lens, is the image real or virtual?
To determine the location and nature (real or virtual) of the image formed by a lens, we can use the lens equation:
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]
where:
- f is the focal length of the lens,
- d_o is the object distance (distance of the object from the lens), and
- d_i is the image distance (distance of the image from the lens).
In this case, the object is located 25 cm in front of the lens, so the object distance (d_o) is -25 cm (negative sign is used for objects located on the same side as the incident light). The lens is a +6.00 lens, which means the focal length (f) is +6.00 cm (positive sign is used for converging lenses).
Using the lens equation:
\[ \frac{1}{6.00} = \frac{1}{-25} + \frac{1}{d_i} \]
Simplifying the equation, we get:
\[ \frac{1}{d_i} = \frac{1}{6.00} - \frac{1}{-25} \]
\[ \frac{1}{d_i} = \frac{-25}{150} - \frac{6.00}{150} \]
\[ \frac{1}{d_i} = \frac{-31}{150} \]
\[ d_i = \frac{-150}{31} \]
Therefore, the image is located approximately 4.84 cm behind the lens (assuming a rounded answer to two decimal places).
To determine whether the image is real or virtual, we simply need to analyze the sign convention. In this case, the lens is converging (+6.00 lens) and the image distance (d_i) is negative. According to conventions, a negative image distance implies that the image is formed on the same side as the object (virtual image).
Therefore, the image formed by the +6.00 lens is a virtual image located approximately 4.84 cm behind the lens.