Answer:
20.0 cm
Explanation:
Here is the complete question
The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Solution
Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.
Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.
Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m
Now, P' = 1/u + 1/v
1/u = P'- 1/v
1/u = 55.0 D - 1/0.02 m
1/u = 55.0 m⁻¹ - 1/0.02 m
1/u = 55.0 m⁻¹ - 50.0 m⁻¹
1/u = 5.0 m⁻¹
u = 1/5.0 m⁻¹
u = 0.2 m
u = 20 cm
So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.
To find average speed, we divide the distance of travel (in this case, 400 metres) by the time she took, 32 seconds. Therefore: 12.5 seconds is her average speed.
Answer:
a ,b , d ,e
Explanation:
took the k12 quiz and got a 100
Answer:
n = 1.4266
Explanation:
Given that:
refractive index of crystalline slab n = 1.665
let refractive index of fluid is n.
angle of incidence θ₁ = 37.0°
Critical angle 
According to Snell's law of refraction:
At point P ; 
Therefore:
Then maximum value of refractive index n of the fluid is:
n = 1.4266
