Question

consider a image that is located 30 cm in front of a lens. it forms an upright image 7.5 cm from the lens. the illumination is so bright that that a faint inverted image, due to reflection off the front of the lens, is observed at 6.0 cm on the incident side of the lens. the lens is then turned around. then it is observed that the faint, inverted image is now 10 cm on the incident side of the lens. what is the index of refraction of the lens

Answer 1

The index of refraction of the lens is 4/15.

Step-by-step explanation:

To find the index of refraction of the lens, we can use the equation:

n = (d1 - d2) / (D1 - D2)

Where:

• n is the index of refraction of the lens
• d1 is the distance of the faint inverted image on the incident side of the lens after the lens is turned around
• d2 is the distance of the faint inverted image on the incident side of the lens before the lens is turned around
• D1 is the distance of the upright image formed by the lens after the lens is turned around
• D2 is the distance of the upright image formed by the lens before the lens is turned around

Given:

• d1 = 10 cm
• d2 = 6 cm
• D1 = 7.5 cm
• D2 = 30 - 7.5 = 22.5 cm (since the original distance of the upright image before turning the lens around is 30 cm)

Substituting the values into the equation, we have:

n = (10 - 6) / (7.5 - 22.5)

Simplifying, we get:

n = -4 / -15 = 4 / 15

Therefore, the index of refraction of the lens is 4/15.