19.03.2023 - 12:50

You may have polarized sunglasses that eliminate glare by polarizing the light. When light is polarized, all of the waves are traveling in parallel planes. Suppose vertically polarized light with inte

Question:

You may have polarized sunglasses that eliminate glare by polarizing the light. When light is polarized, all of the waves are traveling in parallel planes. Suppose vertically polarized light with intensity {eq}I_0 {/eq} strikes a polarized filter with its axis at an angle of {eq}theta {/eq}with the vertical. The intensity of the transmitted light {eq}I_t {/eq} and {eq}theta {/eq} are related by the equation {eq}cos \theta =\sqrt{d\frac{I_t}{I_0}} {/eq}. If {eq}theta = 45^circ {/eq}, write {eq}I_t {/eq} as a function of {eq}I_0 {/eq}.

Answers (1)
  • Gambino
    April 2, 2023 в 16:03

    According to the given equation:

    {eq}cos \theta =\sqrt{\frac{I_t}{I_0}} {/eq}

    When {eq}theta = 45^circ {/eq}, we have:

    {eq}cos 45^circ = \frac{1}{sqrt{2}} {/eq}

    Substituting this value into the equation above:

    {eq}\frac{1}{sqrt{2}} =\sqrt{\frac{I_t}{I_0}} {/eq}

    Squaring both sides:

    {eq}\frac{1}{2} = \frac{I_t}{I_0} {/eq}

    Multiplying both sides by {eq}I_0 {/eq}:

    {eq}I_t = \frac{1}{2} I_0 {/eq}

    Therefore, the intensity of the transmitted light {eq}I_t {/eq} when {eq}theta = 45^circ {/eq} is half of the intensity of the incident light {eq}I_0 {/eq}.

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